Hierarchical Polygamy Inequality for Entanglement of Tsallis q-Entropy?

Yu Luo(羅宇)and Yong-Ming Li(李永明)

1College of Computer Science,Shaanxi Normal University,Xi’an 710062,China

2Centre for Quantum Software and Information(UTS:Q|SI?),Faculty of Engineering and Information Technology,University of Technology Sydney,Australia

1 Introduction

Multipartite entanglement plays a crucial role in quantum physics and is the key resource in many quantum information processing tasks.One of most surprising phenomena for multipartite entanglement is the monogamy property.A simple example of monogamy property can be interpreted as the amount of entanglement between A and B,plus the amount of entanglement between A and C,cannot be greater than the amount of entanglement between A and the pair BC.Monogamy property has been considered in many areas of physics,as like in quantum cryptography,[1?2]condensed matter physics[3?4]and black-hole physics.[5?6]Co ff man et al. first considered three qubits A,B and C which may be entangled with each other,[7]who showed that the squared concurrence C2follows this monogamy inequality.Later,Osborne et al.generalised Co ff man et al.’s result in multi-qubit system.[1]different kinds of monogamy inequalities have also been noted in Refs.[8–33]

As dual to monogamy property,polygamy property in multi-particle systems has arised many interests by researchers.[34?37]Polygamy property was first provided by using the concurrence of assistance to quantify the distributed bipartite entanglement in multi-qubit systems.[34,36]Polygamy property has also been considered in many entanglement measures,such as R′enyi αentropy[26]and Tsallis q-entropy.[37]

Tsallis q-entropy is an important entropic measure,which can be used in many areas of quantum information theory.[38?43]In this paper,we study the polygamy inequality of quantum entanglement in terms of Tsallis qentropy.We first give a lower bound of TOA in the 2?d systems.The lower bound between TEE and TOA is also given in the 2?d system.Furthermore,we prove TOA follows a hierarchical polygamy inequality in a 2?2?2N?2systems.

This paper is organized as follows.In Sec.2,we recall some basic concepts to be used in this paper.In Sec.3,we present our main results.Finally,we summarize our results in Sec.4.

2 Some Basic Concepts

2.1 Concurrence and Concurrence of Assistance

Quantifying entanglement is a core problem in quantum information theory.Given any pure state|ψ?ABin the Hilbert space HA?HB,the concurrence is definedas:[44]

where ρA=TrB(|ψ?AB?ψ|).Note that

with ρB=TrA(|ψ?AB?ψ|).

Given a mixed state ρAB,the concurrence can be defined via the convex-roof extension:[45]

where the minimum is taken over all possible pure state decompositions{pi,|ψi?AB}of ρABwith ∑ipi=1 and pi≥0.

As a dual quantity to concurrence,the concurrence of assistance(COA)can be defined as:

where the maximum is taken over all possible pure state decompositions{pi,|ψi?AB}of ρABwith ∑ipi=1 and pi≥0.

To understand COA better,consider a tripartite pure state|ψ?ABCshared among three parties referred to as Alice,Bob,and Charlie.[34?35]The entanglement supplier,Charlie,performs a measurement on his share of the tripartite state,which yields a known bipartite entangled state for Alice and Bob.Tracing over Charlie’s system yields the bipartite mixed state ρAB=TrC(|ψ?ABC?ψ|)shared by Alice and Bob.Charlie’s aim is to maximize entanglement for Alice and Bob,and the maximum average entanglement he can create is the COA.

For a two-qubit mixed state ρAB,concurrence and COA are known to have analytic formula:[34,44]

where λibeing the eigenvalues,in decreasing order,of matrix

2.2 TEE and TOA

Given a bipartite state ρABin the Hilbert space HA?HB.The Tsallis q-entropy is defined as:[46]

for any q>0 and q1.When q tends to 1,the Tsallis qentropy Tq(ρ)converges to its von Neumann entropy:[47]limq→1Tq(ρ)= ? Tr(ρlnρ).For any pure state|ψAB?,the TEE is defined as:

for any q>0.For a mixed state ρAB,the TEE can be defined as

for any q>0,where the minimum is taken over all possible pure state decompositions{pi,|ψi?AB}of ρAB.TEE can be viewed as a general entanglement of formation when q tends to 1.The entanglement of formation is defined as:[48?49]

whereis the von Neumann entropy,the minimum is taken over all possible pure state decompositions{pi,|ψi?AB}of ρAB.For a mixed state ρAB,the TOA can be defined as:

for any q>0,where the maximum is taken over all possible pure state decompositions{pi,|ψi?AB}of ρAB.

In Ref.[44],Wootters derived an analytical formula for a two-qubit mixed state ρAB:

where H(x)=?xlnx?(1?x)ln(1?x)is the binary entropy and CAB=max{0,λ1?λ2?λ3?λ4}is the concurrence of ρAB,with λibeing the eigenvalues,in decreasing order,of matrix

Tq(ρAB)has an analytical formula for a two-qubit mixed state,which can be expressed as a function of the squared concurrence

where the function fq(x)has the form:

2.3 Three Tangle

For any tripartite pure state|ψ?ABCin a 2 ? 2 ? d system,the three tangle of it is defined as:

For a mixed state ρABC,three tangle can be defined as:

where the minimum is taken over all possible pure state decompositions{pi,|ψi?ABC}of ρABC.

3 Main Results

We will show our main results in this section.

Theorem 1 For any bipartite mixed state ρABin a 2?d system,we have

whereand=Ca(ρAB)is the COA of ρAB.

Proof Let{pi,|ψi?AB}be an optimal convex decomposition for the COA CaAB,then we have

where the second inequality holds is due to Tq(|ψ?AB) ≥fq[C2(|ψ?AB)]for q>0,[33]and we have used the convex-in the third inequality.[33]

Thus,the proof is completed. ?

Equation(16)provides a lower bound for TOA in the 2?d system.

Example 1 Consider the reduced states ρABand ρACof general W state|W?ABC= α|100?+β|010?+γ|001?.From Eq.(5),we have Ca(ρAB)=2|αβ|and Ca(ρAC)=2|αγ|,thus we get the lower bounds:

Now we will study the relationship between TEE and TOA.We have the following theorem first.

Theorem 2 For any tripartite mixed state ρABCin a 2?2?d system,we have

whereand τ3(ρABC)is three tangle of ρABC.

Proof Let us consider a pure state|ψ?ABCfirst.For the state|ψ?ABC,the following inequality holds for q ∈

where we have used the convexity of(x)in the second inequality,[33]the last equality holds because we have equalityfor any tripartite pure state|ψ?ABCin a 2?2?d system.[50]

For a tripartite mixed state ρABCin a 2 ? 2 ? d system,supposeis the optimal decomposition in the senseand letwe have

where the first inequality holds is due to fq(x)is an increase monotonic function of x,and we have used Cauchy-Schwarz inequality

by settingThe second inequality holds is due to the convexity of(x).We have used definition in the last inequality.

Thus,the proof is completed. ?

As an application,we could calculate TOA of some state.

Example 2 Consideratwo-qubitstateρAB=it is easy to show that the threequbit GHZ statepuri fi cation of ρAB.And we have τ(|GHZ?ABC)=1 and Tq(ρAB)=0.From Theorem 2,we have

On theotherhand,forany two-qubitstateρAB,Note that

Thus,we have

which implies

where

Theorem 3 Shows that there is a gap between squared TOA and squared TEE in the 2?2?d system,the gap is connected with three-tangle.Similarly,we have following results for the gap between TOA and TEE:

Theorem 3 For any tripartite mixed state ρABCin a 2?2?d system,we have

where q∈[2,3].

Proof For a tripartite mixed state ρABCin a 2 ? 2 ? 2 system,supposeis the optimal decomposition in the sense ofand let

where the first inequality is due to fq(x)is an increase monotonic function of x,and we have used Cauchy-Schwarz inequality

by settingSecond inequality is due to fq(x?y)≤fq(x)?fq(y)for q∈[2,3],and we have used definition in the last inequality.

Thus,the proof is completed. ?

We also find a relationship between TEE and TOA:

Theorem 4 For any tripartite pure state|ψ>ABCin a 2?2?d system,we have

where

Proof

where we use thein the last equality,and the first inequality holds is due to the convexity of increasing fq(x).

Thus,the proof is completed. ?

The bound in Eq.(19)can be saturated for any tripartite pure statein a 2 ? 2 ? d system.It is obvious that the reduced state ρABis separable,and thus the reduced state ρACis a pure state,which implies

Finally,we obtain a polygamy relation for TOA:

Theorem 5 For any mixed state ρABCin a 2?2?2N?2system,the following hierarchical polygamy relation holds

where

Proof First,we should consider a pure statewe have

where the first inequality is due to

holds in the 2?2?2N?2system,[53]the second inequality is due to Theorem 1.

Second,suppose that the optimal decomposition forwe can derive

Thus,the proof is completed. ?

A straightforward corollary is for any N-qubit mixed state ρA1|A2···AN,the following polygamy inequality holds:

whereThis inequality has been discussed in Ref.[51].We also note that in Ref.[37],Kim derives a sufficient condition for the general polygamy inequality of multipartite quantum entanglement in arbitrary dimensions using Tsallis q-entropy for q≥1(Theorem 1 in his paper).While,the sufficient condition is not easy to derive a certain polygamy inequality because of the hard analysis of parameter q.Our method not only derived a polygamy inequality for the parameter q in a 2?2?2N?2systems,but also the new polygamy inequality holds forThe new polygamy inequality can be seen as a supplement for Kim’s result.

Moreover,we could compare Theorem 5 with our another result in Ref.[54].The main result in Ref.[54]claimed for any states ρABC,TOA satisfies:

This result describes another distribution in the multipartite quantum system than the polygamy relations in Eq.(22),and the equality holds above for arbitrary dimensional space of states.

Example 3 Consider a three-qubit W state|W>ABC=The TOA of|W>A|BCisOn the other hand,from Example 1,we have the lower bound:fq(8/9),we have?.As shown in Fig.1,we plot the function? with?is nonnegative forThus

Fig.1 (Color online)The function?with q∈?is nonnegative for q∈

4 Conclusion

We have provided a one-parameter class of polygamy inequalities in terms of Tsallis q-entropy.We have found a lower bound of TOA in the 2?d systems.The lower bound between TEE and TOA is also given in the 2?d system.Furthermore,we have proven TOA follows a hierarchical polygamy inequality in a 2?2?2N?2systems.A straightforward corollary of this hierarchical polygamy inequality is for any N-qubit mixed state ρA1A2···ANthe general polygamy inequality holds.Based on the one-parameter class of entanglement measurements,some interesting results have been provided in this paper.We hope our results can be a useful tool to understand the property of multi-party quantum entanglement.

Acknowledgments

Y.Luo thanks prof Min-Hsiu Hsieh for comments.

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