Lower Bound of Concurrence Based on Positive Maps
aa r X i v : . [ qu a n t - ph ] A p r Lower Bound of Concurrence Based on Positive Maps
Xiao-Sheng Li , Xiu-Hong Gao , and Shao-Ming Fei , Department of Mathematics, School of Science,South China University of Technology, Guangzhou 510640, China School of Mathematical Sciences, Capital Normal University, Beijing 100048, China Max-Planck-Institute for Mathematics in the Sciences, 04103 Leipzig, Germany
Abstract
We study the concurrence of arbitrary dimensional bipartite quantum systems. An explicitanalytical lower bound of concurrence is obtained, which detects entanglement for some quantumstates better than some well-known separability criteria, and improves the lower bounds such asfrom the PPT, realignment criteria and the Breuer’s entanglement witness.
PACS numbers: 03.67.Mn, 03.67.-a, 02.20.Hj, 03.65.-w H and H be N -dimensional vector spaces. A bipartite quantum pure state | ψ i in2 ⊗ H has a Schmidt form | ψ i = X i α i | e i i ⊗ | e i i , (1)where | e i i and | e i i are the orthonormal bases in H and H respectively, α i are the Schmidtcoefficients satisfying P i α i = 1.The concurrence of the state | ψ i is given by C ( | ψ i ) = q − T rρ ) = 2 sX i 1, where || · || stands for the trace norm of a matrix, T the partial transposition associated with the space H and ˜ ρ the realigned matrix of ρ . While the lower bound in [12] corresponds to a convexfunctional f W ( ρ ) = − T r ( W ρ ), where W is the entanglement witness introduced in [12].3et Φ be a matrix map that maps an N × N matrix A , ( A ) ij = a ij , i, j = 1 , ...N , to an N × N matrix Φ( A ) with (Φ( A )) ij = − a ij for i = j , and (Φ( A )) ii = ( N − a ii + a i ′ i ′ , where i ′ = i + 1 mod N . It can be shown that the matrix map Φ is positive but not completelypositive [19]. Theorem For any bipartite quantum state ρ = P i p i | ψ i ih ψ i | ∈ H ⊗ H , the concurrence C ( ρ ) satisfies C ( ρ ) ≥ s N ( N − 1) ( k ( I N ⊗ Φ) ρ k − ( N − , (6)where I N is the N × N identity matrix. Proof Set f ( | ψ ih ψ | ) = k ( I N ⊗ Φ) | ψ ih ψ |k − ( N − f ( | ψ ih ψ | ) is convex asthe trace norm is convex. What we need to prove is that, for any pure state in Schmidtform (1), the inequality (4) holds.Since the trace norm does change under local coordinate transformation, we take | ψ i =( α , , · · · , , , α , · · · , , , , α , · · · , , · · · · · · , , · · · , , α N ) t , where t denotes transposi-tion, the Schmidt coefficients satisfy 0 ≤ α , α , α , · · · , α N ≤ α + α + α + · · · + α N = 1.It is direct to verify that the matrix T ≡ ( I N ⊗ Φ)( | ψ ih ψ | ) has N − N singular values0, N singular values α , α , α , · · · , α N , the remaining N ones are the singular values of thefollowing matrix B : B = ( N − α − α α − α α · · · − α α N − α α ( N − α − α α · · · − α α N ... ... ... · · · ... − α α N − α α N − α α N · · · ( N − α N . As B is Hermitian and real, its singular values are simply given by the square roots of theeigenvalues of B . In fact we only need to consider the absolute values of the eigenvalues of B . The eigenpolynomial equation of B is: H ( x ) = | xI N − B | = x N − ( N − x N − + ( N − N − X i 0. Therefore H ( x ) is a monotonicallydecreasing function when x < 0. Taking into account that H (0) = 0, we see that there existno negative roots of (7) in this case.The inequality (9) that needs to be proved has the form now, N X i =1 x i − ( N − ≤ X i 0. There are alsono negative roots of (7). One can similarly prove the inequality (9).(b) When β = 0, we have H (0) = ( − N +1 ( N − N − ( α α α · · · α N ). If N is even, wehave H (0) < 0. From (8) we get x x x · · · x N < 0. Therefore, there exists at least onenegative root, say x < 0, such that H ( x ) = 0.Due to that H ′ ( x ) < x < H ( x ) is a monotonically decreasing function when x < 0. Taking into account that H (0) < 0, we have that x < N X i =2 x i − x − ( N − ≤ X i 0, where in thelast step the property of the diagonally dominant matrix ( P i 0, we have that x ≥ − P i 0. The theoremcan be similarly proved.Our bound (6) can detect better entanglement than other lower bounds of concurrence.As an example let us consider a state in 4 × ρ = (1 / diag ( q , q , q , q , q , q , q , q , q , q , q , q , q , q , q , q ) + q i = j X i,j =1 , , , F i,j , (13)where F i,j is the unit matrix with ( i, j )-entry 1 and others 0, q m ≥ P q m = 1, m = 1 , , , N = 4, the positive map Φ maps a matrix M with ( M ) ij = ( m ij ), i, j = 1 , ..., 4, toΦ( M ) = m + m − m − m − m − m m + m − m − m − m − m m + m − m − m − m − m m + m . By direct computation we have the following set of eigenvalues of ( I ⊗ Φ)( ρ ):14 { q + 2 q , q + 2 q , q + 2 q , q + 2 q , q + 2 q , q + 2 q , q + 2 q , q + 2 q ,q + 2 q , q + 2 q , q + 2 q , q + 2 q , q − q , q + q , q + q , q + q } . Therefore from (6) we have C ( ρ ) ≥ r 16 ( k ( I ⊗ Φ) ρ k − 3) = 14 √ q − q + | q − q | ) . (14)From [11], with respect to the PPT and realignment operation one has bounds C P P T ( ρ ) ≥ q ( k ρ T k − √ (2 q + | q − q | + (cid:12)(cid:12)(cid:12) q + q − p q + ( q − q ) (cid:12)(cid:12)(cid:12) + p q + ( q − q ) − 1) (15)6nd C r ( ρ ) ≥ q ( k ˜ ρ k − q (cid:16) q + ( p ( q − q + q − q ) + 2 p ( q − q ) + ( q − q ) − (cid:17) . (16)From the formula presented in [12], one has the bound C W ( ρ ) ≥ r 16 ( − tr ( W ρ )) = − √ q + 2 q + q ) . (17) Θ FIG. 1: Lower bounds for the state (13). Solid line: the lower bound given by (14); Dashed line:lower bound given by (15); Dashed-dotted line: lower bound given by (16); θ axis: lower boundgiven by (17). To compare among these bounds, let us take q = , q = 0 . q = (1 − q − q ) sin θ , q = (1 − q − q ) cos θ , θ ∈ [0 , π ]. From Fig. 1 we see that our new bound (14) detectsentanglement for θ > . 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