aa r X i v : . [ qu a n t - ph ] D ec Monogamy Properties of Qubit Systems
Xue-Na Zhu and Shao-Ming Fei , School of Mathematics and Statistics Science, Ludong University, Yantai 264025, China School of Mathematical Sciences, Capital Normal University, Beijing 100048, China Max-Planck-Institute for Mathematics in the Sciences, 04103 Leipzig, Germany
We investigate monogamy relations related to quantum entanglement for n − qubit quantum sys-tems. General monogamy inequalities are presented to the β th ( β ∈ (0 , β th ( β ∈ (0 , √ β . In additions, new monogamy relations are also derived whichinclude the existing ones as special cases. PACS numbers:
I. INTRODUCTION
Quantum entanglement [1–5] lies at the heart of quantum information processing and quantum computation[6].Accordingly its quantification has drawn much attention in the last decade. As one of the fundamental differencesbetween quantum entanglement and classical correlations, a key property of entanglement is that a quantum systementangled with one of other systems limits its entanglement with the remaining systems. The monogamy relationsgive rise to the structures of entanglement distribution in multipartite systems. Monogamy is also an essential featureallowing for security in quantum key distribution [8].For a tripartite system A , B and C , the monogamy of an entanglement measure ε implies that [9], the entanglementbetween A and BC satisfies ε A | BC ≥ ε AB + ε AC . Such monogamy relations are not always satisfied by any entanglementmeasures. It has been shown that the squared concurrence C [10, 11] and the squared entanglement of formation E [12, 13] do satisfy such monogamy relations. In Ref.[14] it has been shown that the general monogamy inequalities aresatisfied by the α ( α ≥ C α and the α ( α ≥ √ E α for n − qubit mixed states. Another useful entanglement measure is the negativity[7], a quantitative version of Peres’scriterion for separability. The authors in Ref.[15] studied the monogamy property of the α th power of negativity N α ( α ≥
2) and discussed tighter α th ( α ≥
2) power of the convex-roof extended negativity (CREN) ˜ N α . In Ref.[16]tighter monogamy inequalities for concurrence, entanglement of formation and CREN has been investigated for α ≥ α th (0 < α <
2) power of concurrence, negativityand CREN, and the α th (0 < α < √
2) power of entanglement of formation. In this paper, we study the generalmonogamy inequalities of C β , N β , ˜ N β and E β for β ∈ [0 , M ], where M is any real number greater than zero. II. MONOGAMY PROPERTY OF CONCURRENCE
For a bipartite pure state | ψ i AB , the concurrence is given by [17–19], C ( | ψ i AB ) = q − T r ( ρ A )] , (1)where ρ A is reduced density matrix by tracing over the subsystem B , ρ A = T r B ( | ψ i AB h ψ | ). The concurrence isextended to mixed states ρ = P i p i | ψ i ih ψ i | , p i ≥ P i p i = 1, by the convex roof construction, C ( ρ AB ) = min { p i , | ψ i i} X i p i C ( | ψ i i ) . (2)For n − qubit quantum states, the concurrence satisfies [14] C αA | B B ...B n − ≥ C αAB + ... + C αAB n − , (3)for α ≥
2, where C A | B B ...B n − is the concurrence of ρ under bipartite partition A | B B ...B n − , and C AB i , i =1 , ..., n −
1, is the concurrence of the mixed states ρ AB i = T r B B ...B i − B i +1 ...B n − ( ρ ). For C AB i = 0, i = 1 , ..., n − C αA | B ...B n − < C αAB + ... + C αAB n − , (4)for α ≤
0. Further, in Ref. [16] tighter monogamy inequalities than (3) are derived for the α th ( α ≥
2) power ofconcurrence.
Lemma 1
For real numbers x ∈ [0 , and t ≥ , we have (1 + t ) x ≥ x − t x . [Proof] Let g x ( t ) = (1+ t ) x − t x with x ∈ [0 ,
1] and t ∈ [1 , + ∞ ). Since dg x ( t ) dt = xt − ( x +1) [1 − (1 + t ) x − ] ≥
0, we obtainthat g x ( t ) is an increasing function of t . Hence, g x ( t ) ≥ g x (1), i.e, (1 + t ) x ≥ x − t x . Theorem 1
For any ⊗ ⊗ n − tripartite mixed state:(1) if C AB ≤ C AC , the concurrence satisfies C βA | BC ≥ C βAB + (2 βα − C βAC , (5) where ≤ β ≤ α and α ≥ .(2) if C AB ≥ C AC , the concurrence satisfies C βA | BC ≥ (2 βα − C βAB + C βAC , (6) where ≤ β ≤ α and α ≥ . [Proof] For arbitrary 2 ⊗ ⊗ n − tripartite state ρ ABC , one has [14], C αA | BC ≥ C αAB + C αAC . If max { C AB , C AC } = 0, i.e., C AB = C AC = 0, obviously we have the inequalities (5) or (6); If min { C AB , C AC } = 0,obviously C βA | BC ≥ max { C βAB , C βAC } ≥ (2 βα −
1) max { C βAB , C βAC } with 0 ≤ β ≤ α , we also have the inequalities (5) or(6).If max { C AB , C AC } > { C AB , C AC } 6 = 0, assuming 0 < C AB ≤ C AC , we have C αxA | BC ≥ ( C αAB + C αAC ) x = C αxAB (cid:16) C αAC C αAB (cid:17) x ≥ C αxAB (cid:18) x − (cid:16) C αAC C αAB (cid:17) x (cid:19) = C αxAB + (2 x − C αxAC , where the second inequality is due to the inequality (1 + t ) x ≥ x − t x for 0 ≤ x ≤ t = C αAC C αAB ≥
1. Denote αx = β . Then β ∈ [0 , α ] since x ∈ [0 ,
1] and one gets the inequality (5). If C AB ≥ C AC , similar proof gives theinequality (6).One can see that Theorem 1 reduces to the monogamy inequality (3) if β = α ≥
2. In particular, if we take β = 1, we have C A | BC ≥ min { C AB , C AC } + (2 α −
1) max { C AB , C AC } for α ≥
2. And the tighter relation is C A | BC ≥ min { C AB , C AC } + ( √ −
1) max { C AB , C AC } . Example 1.
Let us consider the three-qubit case. Any three-qubit state | ψ i can be written in the generalizedSchmidt decomposition [14, 20, 21], | ψ i = λ | i + λ e iϕ | i + λ | i + λ | i + λ | i , (7)where λ i ≥ i = 0 , ...,
4, and P i =0 λ i = 1. From Eq.(1) and Eq.(2), we have C A | BC = 2 λ p λ + λ + λ , C AB =2 λ λ , and C AC = 2 λ λ . Without loss of generality, we set λ = cos θ , λ = sin θ cos θ , λ = sin θ sin θ cos θ ,λ = sin θ sin θ sin θ cos θ , and λ = sin θ sin θ sin θ sin θ , θ i ∈ [0 , π ]. Assume λ ≥ λ , i.e C AC ≥ C AB :(a) if θ = π , we have C βA | BC − C βAB − (2 βα − C βAC = (2 λ ) β h ( λ + λ + λ ) β − λ β − (2 βα − λ β i = (2 λ ) β sin β θ sin β θ h − (2 βα −
1) cos β θ i ≥ (2 λ ) β sin β θ sin β θ (2 − βα ) ≥ , where 0 ≤ β ≤ α , α ≥ cosθ ≤ θ = π , we denote t = sin α θ cos α θ cos α θ . We have C βA | BC − C βAB − (2 βα − C βAC = (2 λ ) β h ( λ + λ + λ ) β − λ β − (2 βα − λ β i = (2 λ ) β sin β θ sin β θ h − cos β θ − (2 βα −
1) sin β θ cos β θ i = (2 λ ) β sin β θ sin β θ (cid:20) − cos β θ (cid:18) βα − t βα (cid:19)(cid:21) ≥ (2 λ ) β sin β θ sin β θ h − cos β θ (1 + t ) βα i = (2 λ ) β sin β θ sin β θ h − ( cos α θ + sin α θ cos α θ ) β i ≥ , where 0 ≤ β ≤ α and α ≥
2. The first inequality is due to Lemma 1 with 0 ≤ x = βα ≤ cos α θ + sin α θ cos α θ ≤ α ≥ C βA | BC ≥ C βAB + (2 βα − C βAC for 0 ≤ β ≤ α and α ≥
2. For the case λ ≤ λ , i.e., C AB ≥ C AC , similarly one obtains that C βA | BC ≥ (2 βα − C βAB + C βAC with 0 ≤ β ≤ α and α ≥ Theorem 2
For any n -qubit quantum state ρ such that C AB i ≤ C A | B i +1 ...B n − for i = 1 , ..., m, and C AB j ≥ C A | B j +1 ...B n − for j = m + 1 , ..., n − , ∀ ≤ m ≤ n − , n ≥ , we have C β ( ρ A | B B ...B n − ) ≥ m X i =1 (2 βα − i − C β ( ρ AB i ) (8)+ (2 βα − m +1 n − X i = m +1 C β ( ρ AB i ) + (2 βα − m C β ( ρ AB n − ) , where ≤ β ≤ α and α ≥ . [Proof] For convenience, we denote r = 2 βα −
1. For any 2 ⊗ ⊗ ⊗ ... ⊗ ρ AB ...B n − , we have C βA | B B ...B n − ( ρ ) ≥ C βAB + rC βA | B ...B n − ≥ C βA | B + rC βA | B + r C βA | B ...B n − ≥ ... ≥ m X i =1 r i − C βAB i + r m C βA | B m +1 ...B n − ≥ m X i =1 r i − C βAB i + r m h rC βAB m +1 + C βA | B m +2 ...B n − i ≥ ... ≥ m X i =1 r i − C βAB i + r m +1 n − X i = m +1 C βAB i + r m C βAB n − , where the first four inequalities are due to C AB i ≤ C A | B i +1 ...B n − ( i = 1 , ..., m ) and the inequality (5), the last threeinequalities are due to C AB j ≥ C A | B j +1 ...B n − ( j = m + 1 , ..., n −
2) and the inequality (6).For an n -qubit quantum state ρ AB ...B n − , in Ref.[14] it has been shown that the β th concurrence C β (0 < β < C β ( | ψ i A | B B ...B N − ) ≥ P n − i =1 C β ( ρ AB i ). Theorem (2) first time givesthe monogamy inequality satisfied by the β − th concurrence C β for the case of (0 < β < β = 1 and α = 2, we get the monogamy relation satisfied by the concurrence C : C ( ρ A | B B ...B N − ) ≥ m X i =1 ( √ − i − C ( ρ AB i ) + ( √ − m +1 n − X i = m +1 C ( ρ AB i ) + ( √ − m C ( ρ AB n − ) . Β Α u FIG. 1: u ( β, α ) for 0 ≤ β ≤ α ≥ Example 2.
Let us consider the pure state | ψ i (7) in the Example 1. Set λ = √ , λ = , λ = , λ = √ and λ = √ . We have C A | BC = √ ≈ .
707 and C AB + C AC = √ √ ≈ . . One can see that C A | BC < C AB + C AC .Denoting u ( β, α ) = C βA | BC − C βAB − (2 βα − C βAC = ( √ ) β − ( √ ) β − (2 α − √ ) β with 0 ≤ β ≤ α and α ≥ , wehave u (1 , α ) ≥ .
201 for all α ≥
2. Furthermore, our result shows that u ( β, α ) ≥ ≤ β ≤ α ≥
2, seeFig. 1.
III. MONOGAMY INEQUALITY FOR NEGATIVITY
Given a bipartite state ρ AB , the negativity is defined by [24] N ( ρ AB ) = || ρ T A AB || − , where ρ T A AB is the partially transposed matrix of ρ AB with respect to the subsystem A , || X || = T r √ XX † denotes thetrace norm of X . For the convenience of discussion, we use the following definition of negativity: N ( ρ AB ) = || ρ T A AB || − . It has been shown that for any n -qubit pure state | ψ i A | B ...B n − , the negativity satisfies the monogamy inequalityholds for α ≥ N αA | B ...B n − ( | ψ i ) ≥ N αAB + ... + N αAB n − , and the polygamy inequality for α ≤ N αA | B ...B n − ( | ψ i ) < N αAB + ... + N αAB n − . Here N A | B ...B n − ( | ψ i ) is the negativity of | ψ i under bipartite partition A | B ...B n − , and N AB i is the negativity ofthe quantum state ρ AB i = T r B ...B i − B i +1 ...B n − ( | ψ i ). In the following we study the monogamy property of the β thpower of negativity N β for β ∈ (0 , Theorem 3
For any n -qubit quantum pure state | ψ i such that C AB i ≤ C A | B i +1 ...B n − for i = 1 , ..., m , and C AB j ≥ C A | B j +1 ...B n − for j = m + 1 , ..., n − , ∀ ≤ m ≤ n − and n ≥ , we have N β ( | ψ i A | B B ...B N − ) ≥ m X i =1 (2 βα − i − N β ( ρ AB i ) (9)+ (2 βα − m +1 n − X i = m +1 N β ( ρ AB i ) + (2 βα − m N β ( ρ AB n − ) , where ≤ β ≤ α and α ≥ . Theorem 3 can be seen by using (8) in theorem 2, and noting that C ( | ψ i A | BC = N ( | ψ i A | BC ) for 2 ⊗ t ⊗ s ( t ≥ , s ≥ N ( ρ AB ) ≤ C ( ρ AB ) for 2 ⊗ m systems.Given a bipartite state ρ AB , the CREN is defined as the convex roof extended negativity of pure states [15, 22]˜ N ( ρ AB ) = min { p i , | ψ i i} X i p i N ( | ψ i i ) , with the infimum taking over all possible decompositions of ρ AB in a mixture of pure states, ρ AB = P i p i | ψ i ih ψ i | , p i ≥ P i p i = 1.For a mixed state ρ ABC in 2 ⊗ ⊗ n − systems, the following monogamy inequality holds [15], ˜ N αA | BC ( ρ ) ≥ ˜ N αAB + ˜ N αAC for α ≥
2, and the following polygamy inequality holds, ˜ N αA | BC ( ρ ) < ˜ N αAB + ˜ N αAC for α ≤
0. Formultiqubit mixed states ρ AB ...B n − , one has the following monogamy inequality for the α th power of CREN for α ≥ N αA | B ...B n − ( ρ ) ≥ ˜ N αAB + ... + ˜ N αAB n − , and the following polygamy inequality for α ≤ N αA | B ...B n − ( ρ ) < ˜ N αAB + ... + ˜ N αAB ...B n − , where ˜ N A | B ...B n − ( ρ ) is the negativity of ρ under bipartite partition A | B ...B n − , and ˜ N AB i is the negativity of thequantum state ρ AB i = T r B ...B i − B i +1 ...B n − ( ρ ).With a similar consideration to concurrence, we obtain the following result. Corollary 1
For any ⊗ ⊗ n − mixed state ρ ABC , and ≤ β ≤ α , α ≥ :(1) if ˜ N AB ≤ ˜ N AC , the CREN satisfies ˜ N βA | BC ≥ ˜ N βAB + (2 βα −
1) ˜ N βAC ; (10) (2) if ˜ N AB ≥ ˜ N AC , the CREN satisfies ˜ N βA | BC ≥ (2 βα −
1) ˜ N βAB + ˜ N βAC . (11) Corollary 2
For any n -qubit quantum state ρ AB ...B n − such that ˜ N AB i ≤ ˜ N A | B i +1 ...B n − ( i = 1 , ..., m ) and ˜ N AB j ≥ ˜ N A | B j +1 ...B n − ( j = m + 1 , ..., n − , ∀ ≤ m ≤ n − , n ≥ , we have ˜ N β ( ρ A | B B ...B N − ) ≥ m X i =1 (2 βα − i − ˜ N β ( ρ AB i ) + (2 βα − m +1 n − X i = m +1 ˜ N β ( ρ AB i ) + (2 βα − m ˜ N β ( ρ AB n − ) , where ≤ β ≤ α and α ≥ . IV. MONOGAMY INEQUALITY FOR EOF
The entanglement of formation (EoF) [23, 24] is a well-defined and important measure of quantum entanglementfor bipartite systems. Let H A and H B be m - and n -dimensional ( m ≤ n ) vector spaces, respectively. The EoF of apure state | ψ i ∈ H A ⊗ H B is defined by E ( | ψ i ) = S ( ρ A ), where ρ A = T r B ( | ψ ih ψ | ) and S ( ρ ) = T r ( ρ log ρ ). For abipartite mixed state ρ AB ∈ H A ⊗ H B , the entanglement of formation is given by E ( ρ AB ) = min { p i , | ψ i i} X i p i E ( | ψ i i ) , with the infimum taking over all possible decompositions of ρ AB in a mixture of pure states ρ AB = P i p i | ψ i ih ψ i | ,where p i ≥ P i p i = 1.Denote f ( x ) = H (cid:16) √ − x (cid:17) , where H ( x ) = − x log ( x ) − (1 − x ) log (1 − x ) . One has [16] E ( ρ AB ) ≥ f ( C AB ) . (12) Lemma 2 If ≤ x ≤ y ≤ , we have f β ( x + y ) ≥ f β ( x ) + (2 βα − f β ( y ) , (13) where f β ( x + y ) = ( f ( x + y )) β , ≤ β ≤ α and α ≥ √ . [Proof] Since 0 ≤ x ≤ y ≤ f ( x ) is a monotonically increasing function for 0 ≤ x ≤
1, one has f ( x ) ≤ f ( y )and f α ( x + y ) ≥ f α ( x ) + f α ( y ) for α ≥ √ z ∈ [0 , x = 0, i.e., f ( x ) = 0, we have f αz ( x + y ) ≥ ( f α ( x ) + f α ( y )) z = ( f α ( y )) z ≥ (2 z − f αz ( y )) . If x = 0, i.e., f ( x ) = 0, we have f αz ( x + y ) ≥ ( f α ( x ) + f α ( y )) z = f αz ( x ) (cid:18) f α ( y ) f α ( x ) (cid:19) z ≥ f αz ( x ) + (2 z − f αz ( y )) , where the last inequality is obtained by using lemma 1. The Lemma 2 is proved by setting αz = β .It has been shown that the entanglement of formation does not satisfy monogamy inequality such as E AB + E AC ≤ E A | BC [13]. In [14] the authors showed that E α ( ρ A | B B ...B n − ) ≥ P n − i =1 E α ( ρ AB i ) for α ≥ √
2, and E α ( ρ A | B B ...B n − ) ≤ P n − i =1 E α ( ρ AB i ) for α ≤
0. In Ref. [16] tighter monogamy relation for E α ( α ≥ √
2) hasbeen derived for n -qubit states.In fact, applying the same approach to the theorems 1 and 2, we can prove the following results generally: Theorem 4
For any ⊗ ⊗ mixed state ρ ∈ H A ⊗ H B ⊗ H C , and ≤ β ≤ α , α ≥ √ .(1) If C AB ≤ C AC , we have E βA | BC ≥ E βAB + (2 βα − E βAC ; (14) (2) If C AB ≥ C AC , we have E βA | BC ≥ (2 βα − E βAB + C βAC . (15) [Proof] Let α ≥ √ β ∈ [0 , α ]. If C AB ≤ C AC , we have E βA | BC ≥ f β ( C A | BC ) ≥ f β ( C AB + C AC ) ≥ f β ( C AB ) + (2 βα − f β ( C AC )= E βAB + (2 βα − E β AC , where the first inequality is due to the inequality (12), the second inequality is obtained from the inequality C A | BC ≥ C AB + C AC , the third inequality holds because of Lemma 2 and the last equality is obtained from E ( ρ ) = f ( C ( ρ ))for two qubit states. The result for the case 2 can be similarly proved. Example 3.
Consider the W state, | W i = √ ( | i + | i + | i ). We have E A | BC = 0 . E AB = E AC = 0 . E A | BC < E AB + E AC . It is easily verified that E A | BC > . α ≥√ (cid:16) E AB + (2 α − E AC (cid:17) . Denote u ( β, α ) = E βA | BC − E βAB − (2 βα − E βAC = 0 . β − βα × . β .For 0 ≤ β ≤ α ≥ √
2, we have u ( β, α ) ≥
0, see Fig.2.For n − qubit quantum states, we have follow theorem. Β Α u FIG. 2: u ( β, α ) for 0 ≤ β ≤ α ≥ √ Theorem 5
For any n − qubit mixed state ρ AB ...B n − such that C AB i ≤ C A | B i +1 ...B n − ( i = 1 , ..., m ) and C AB j ≥ C A | B j +1 ...B n − ( j = m + 1 , ..., n − , ∀ ≤ m ≤ n − and n ≥ , we have E βA | B B ...B N − ( ρ ) ≥ m X i =1 (2 βα − i − E β ( ρ AB i ) + (2 βα − m +1 n − X i = m +1 E β ( ρ AB i ) + (2 βα − m E β ( ρ AB n − ) , where ≤ β ≤ α and α ≥ √ , E A | B B ...B n − is the entanglement of formation of ρ under bipartite parti-tion A | B B ...B n − , and E AB i , i = 1 , ..., n − , is the entanglement of formation of the mixed state ρ AB i = T r B B ...B i − B i +1 ...B n − ( ρ ) . [Proof] Denote k = 2 βα −
1. For α ≥ √ β ∈ [0 , α ], we have E βA | B B ...B N − ≥ f β ( C A | B B ...B N − ) (16) ≥ f β ( C AB + C A | B ...B n − ) ≥ f β ( C AB ) + kf β ( C A | B ...B n − ) ≥ ... ≥ m X i =1 k i − f β ( C AB i ) + k m f β ( C A | B m +1 ...B n − )= m X i =1 k i − E β ( ρ AB i ) + k m f β ( C A | B m +1 ...B n − ) , where the first inequality is due to (12), the third to the fifth inequalities are due to C A | B i ≤ C A | B i +1 ...B n − ( i =1 , ..., m ) and Lemma 2. Moreover, f β ( C A | B m +1 ...B n − ) ≥ f β ( C AB m +1 + C A | B m +2 ...B n − ) (17) ≥ kf β ( C A | B m +1 ) + f β ( C A | B m +2 ...B n − ) ≥ ... ≥ k n − X i = m +1 f β ( C A | B i ) + f β ( C AB n − )= k n − X i = m +1 E β ( ρ AB i ) + E β ( ρ AB n − ) , where the second to the fourth inequalities are due to C AB i ≥ C A | B i +1 ...B n − ( i = m + 1 , ..., n −
2) and Lemma 2.Combining (16)and (17) we obtain the theorem 5.Theorem 5 gives the monogamy relations satisfied by the β th (0 ≤ β ≤ α , α ≥ √
2) power of EoF for n -qubit states,which is a problem remained unsolved in Ref.[14, 16] for β ∈ (0 , √ β = α ≥ √ , Theorem 5 reduces tothe result in Ref.[14]. In addition if we take β = 1 and α = √ E ( | ψ i A | B B ...B n − ) ≥ m X i =1 (2 √ − i − E ( ρ AB i ) + (2 √ − m +1 n − X i = m +1 E ( ρ AB i ) + (2 √ − m E ( ρ AB n − ) , which gives first time the tight monogamy inequality satisfied by the entanglement of formation itself. V. CONCLUSION
Entanglement monogamy is a fundamental property of multipartite entangled states. We have investigated themonogamy relations related to the concurrence, the negativity, CREN and the entanglement of formation for general n -qubit states. We have derived the monogamy inequalities satisfied by C β , N β , ˜ N β for β ∈ (0 , E β for β ∈ (0 , √
2) for n -qubit states. These monogamy relations are complementary to the existing ones with differentregions of parameter β . Our new monogamy relations also include the existing ones as special cases. Our approachmay be used to study further monogamy properties related to other quantum entanglement measures such as Tsallis- q entanglement and to quantum correlations such as quantum discord. Acknowledgments
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