PhysRevA.83.034303

PHYSICAL REVIEW A83,034303(2011)

Lower bound of concurrence based on positive maps

Xiao-Sheng Li,1Xiu-Hong Gao,2and Shao-Ming Fei2,3

1Department of Mathematics,School of Science,

South China University of Technology,Guangzhou510640,China

2School of Mathematical Sciences,Capital Normal University,Beijing100048,China

3Max-Planck-Institute for Mathematics in the Sciences,D-04103Leipzig,Germany

(Received20January2011;published25March2011)

We study the concurrence of arbitrary dimensional bipartite quantum systems.An explicit analytical lower bound of concurrence is obtained which detects entanglement for some quantum states better than some well-known separability criteria and improves the lower bounds,such as from the positive partial transposition, realignment criteria,and Breuer’s entanglement witness.

DOI:10.1103/PhysRevA.83.034303PACS number(s):03.67.Mn,02.20.Hj,03.65.?w

Quantum entangled states are the most important resource in quantum information processing[1].An important theo-retical challenge in the theory of quantum entanglement is to give a proper description and quanti?cation of quantum entanglement for given quantum states.The entanglement of formation(EOF)[2]and concurrence[3]are two well-de?ned quantitative measures of quantum entanglement.For the two-qubit case EOF is a monotonically increasing function of the concurrence,and an elegant formula for the concurrence was derived analytically by Wootters in[4].It plays an essential role in describing quantum phase transitions in various interacting quantum many-body systems[5,6]and can be experimentally measured[7].However,for the general highly dimensional case,due to the extremizations involved in the calculation,only a few explicit analytic formulas for EOF and concurrence have been found for some special symmetric states[8].

Instead of analytic formulas,some progress has been made toward the lower bounds of EOF and concurrence.For instance,in[9]a lower bound of concurrence that can be tightened by numerical optimization over some parameters has been derived.In[10,11]analytic lower bounds on EOF and concurrence for any dimensional mixed bipartite quantum states have been presented by using the positive partial trans-position(PPT)and realignment separability criteria.These bounds are exact for some special classes of states and can be used to detect many bound entangled states.In[12]another lower bound on EOF for even dimensional bipartite states has been presented from a new separability criterion[13].

A lower bound of concurrence based on local uncertainty relations(LURs)criterion is derived in[14].This bound is further optimized in[15].In[16,17]the authors presented lower bounds of concurrence for bipartite systems in terms of a different approach that has a close relationship with the distillability of bipartite quantum states.In[18]experimentally measurable bounds for EOF have been presented.All these bounds obtained so far together give rise to a good quantitative estimation of EOF and concurrence.In particular,they are supplementary in detecting entanglement.

In this Brief Report,based on positive maps,we present a new lower bound of concurrence for arbitrary dimensional bipartite systems.This bound is shown to detect entanglement that cannot be recognized by the bounds in[11,12].

Let H1and H2be N-dimensional vector spaces.A bipartite quantum pure state|ψ in H1?H2has a Schmidt form

|ψ =

i

αi

e1i

?

e2i

,(1)

where|e1

i

and|e2i are the orthonormal bases in H1and H2,respectively,andαi represents the Schmidt coef?cients satisfying

i

α2

i

=1.

The concurrence of the state|ψ is given by

C(|ψ )=

2(1?Trρ2

1

)=2

i

α2

i

α2

j

,(2)

where the reduced density matrixρ1is obtained by tracing over the second subsystem of the density matrixρ=|ψ ψ|,ρ1=Tr2|ψ ψ|.

A general mixed state in H1?H2has pure state decom-positions,ρ=

i

p i|ψi ψi|,where p i 0and

i

p i=1. The concurrence is extended to mixed statesρby the convex roof,

C(ρ)=min

i

p i C(|ψi ),(3)

where the minimum is taken over all possible convex decom-positions ofρinto an ensemble{|ψi }of pure states with probability distribution{p i}.

Let f(ρ)be a real-valued and convex functional on the total state space with the following property,

f(|ψ ψ|) 2

i

αiαj,(4)

for all state vectors|ψ with Schmidt decompositions(1).By using the inequality in[11],

i

α2iα2j

2

N(N?1)

i

αiαj

2

.

Breuer[12]has derived that C(ρ)satis?es

C(ρ)

2

N(N?1)

f(ρ).(5)

The f(ρ)corresponding to the lower bounds in[11]are the ones with respect to the PPT criterion and the realignment cri-terion,f P P T(ρ)=||ρT1||?1,f r(ρ)=||?ρ||?1,where||·||

represents the trace norm of a matrix,T 1the partial trans-position associated with the space H 1,and ?ρ

the realigned matrix of ρ.The lower bound in [12]corresponds to a convex functional f W (ρ)=?Tr(Wρ),where W is the entanglement witness introduced in [12].

Let be a matrix map that maps an N ×N matrix A ,(A )ij =a ij ,i,j =1,...,N ,to an N ×N matrix (A )with [ (A )]ij =?a ij for i =j ,and [ (A )]ii =(N ?2)a ii +a i i ,where i =i +1mod N .It can be shown that the matrix map is positive but not completely positive [19].

Theorem.For any bipartite quantum state ρ= i p i |ψi ψi |∈H 1?H 2,the concurrence C (ρ)satis?es

C (ρ)

2

N (N ?1)

[ (I N ? )ρ ?(N ?1)],

(6)

where I N is the N ×N identity matrix.

Proof.Set f (|ψ ψ|)= (I N ? )|ψ ψ| ?(N ?1).Obviously f (|ψ ψ|)is convex as the trace norm is convex.What we need to prove is that for any pure state in Schmidt form (1),the inequality (4)holds.

Since the trace norm does change under local coordinate transformation,we take |ψ =(α1,0,...,0,0,α2,...,0,0,0,α3,...,0,...,0,...,0,αN )t ,where t denotes transposition,and the Schmidt coef?cients satisfy 0 α1,α2,α3,...,αN 1,α21+α22+α23+···+α2

N =1.

It is direct to verify that the matrix T ≡(I N ? )(|ψ ψ|)has N 2?2N singular values 0,N singular values α21,α22,α23,...,α2N ,and the remaining N values are the singular values of the following matrix B :

B =?

?????(N ?2)α2

1?α1α2

?α1α3...?α1αN

?α1α

2(N ?2)α2

2

?α2α3

···?α2αN ...............

?α1αN

?α2αN ?α3αN

...

(N ?2)α2

N

?

??

???.As B is Hermitian and real,its singular values are simply given by the square roots of the eigenvalues of B 2.In fact we need only consider the absolute values of the eigenvalues of B .The eigenpolynomial equation of B is

H (x )=|xI N ?B |=x N ?(N ?2)x N ?1

+(N ?3)(N ?1)?? i

α2i αj ??2x N ?2

?(N ?4)(N ?1)2

i

α2i α2

j αk

2

x N ?3+···

+(?1)N

(N ?1)

N ?3

i 1

α2i 1α2i 2

...αi N ?2

2

x 2+(?1)N +1(N ?1)N ?1 α21α22α23...α2

N =0.

(7)

Let x 1,x 2,x 3,...,x N denote the N roots of Eq.(7).By using the relations between roots and coef?cients of the polynomial equation,one has

N i =1

x i =N ?2, N i =1x i

=(?1)2N +1(N ?1)N ?1 N i =1α2

i .

(8)

The inequality (4)that needs to be proved now has the

form

N i =1

|x i |?(N ?2) 2 i

αi αj ,

(9)

where N i =1α2

i =1has been taken into account.

To deal with the eigenpolynomial equation (7),we set β= N

i =1α2i .

(a)If β=0,then H (0)=0,where 0is an eigenvalue of B .From the derivative of H (x )with respect to x ,

H (x )=Nx N ?1?(N ?2)(N ?1)x N ?2

+(N ?3)(N ?2)(N ?1)

i

α2i αj 2x N ?3

?...+2(?1)N (N ?1)N ?3

×

i 1

α2i 1α2i 2...α2i N ?2 x,

(10)

we know that if N is even,H (x )<0when x <0.Therefore

H (x )is a monotonically decreasing function when x <0.Taking into account that H (0)=0,we see that there exist no negative roots of (7)in this case.

The inequality (9)that needs to be proved now has the form

N i =1

x i ?(N ?2) 2 i

αi αj .

(11)

According to the relation in (8),the left hand of the inequality

(11)is zero.Hence the inequality (9)is satis?ed.

When N is odd,H (x )is a monotonically increasing function for x <0.There are also no negative roots of (7).One can similarly prove the inequality (9).

(b)When β=0,we have H (0)=(?1)N +1(N ?

1)N ?1(α21α22α23...α2

N ).If N is even,we have H (0)<0.From (8)we get x 1x 2x 3...x N <0.Therefore,there exists at least one negative root,say x 1<0,such that H (x 1)=0.

Because H (x )<0for x <0,H (x )is a monotonically decreasing function when x <0.Taking into account that H (0)<0,we have that x 1<0is the only negative root.Hence

the inequality (9)needing to be proved becomes

N

i =2

x i ?x 1?(N ?2) 2 i

αi αj .

(12)

From Eq.(8)we need only to prove that x 1 ?

i

From the de?nition of H (x ),we have H (?

i

|?(

i

matrix (

i

0 H (?

i

function when x <0,we have that x 1 ?

i

When N is odd,H (x )is a monotonically increasing function when x <0.The theorem can 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 4×4[19

],

ρ=(1/4)diag(q 1,q 4,q 3,q 2,q 2,q 1,q 4,q 3,q 3,q 2,q 1,q 4,q 4,q 3,q 2,q 1)+q 1

4i =j i,j =1,6,11,16

F i,j ,

(13)

where F i,j is the unit matrix with (i,j )entry 1and others 0,q m 0,

q m =1,m =1,2,3,4.For N =4,the positive map maps a matrix M with (M )ij =(m ij ),i,j =1,...,4,to

(M )=?????2m 11+m 22?m 12?m 13?m 14?m 212m 22+m 33?m 23?m 24?m 31?m 322m 33+m 44?m 34?m 41?m 42?m 432m 44+m 11

???

??

.

By direct computation we have the following set of eigenvalues

of (I 4? )(ρ):

1

4

{q 1+2q 2,q 1+2q 2,q 1+2q 2,q 1+2q 2,q 2+2q 3,q 2+2q 3,q 2+2q 3,q 2+2q 3,

q 3+2q 4,q 3+2q 4,q 3+2q 4,q 3+2q 4,q 4?q 1,3q 1+q 4,3q 1+q 4,3q 1+q 4}.Therefore from (6)we have

C (ρ)

16( (I 4? )ρ ?3)=1

4√6

(q 1?q 4+|q 1?q 4|).(14)

0.2

0.4

0.60.8

θ

0.02

0.040.060.08B o u n d

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);θaxis:lower bound given by (17).

From [11],with respect to the PPT and realignment operation one has bounds

C P P T (ρ)

1

6

( ρT 1 ?1)=1

2√6(2q 1+|q 1?q 3|+ q 2+q 4? 4q 21+(q 2?q 4)2

+

4q 2

1+(q 2?q 4)2?1)(15)and

C r (ρ)

1

6( ?ρ ?1)=

16(3q 1+14[ 1?q 2+q 3?q 4+2

(q 1?q 3)2+(q 2?q 4)2

?3]).

(16)

From the formula presented in [12],one has the bound

C W (ρ)

1

6[?tr (Wρ)]=?12√6

(q 2+2q 3+q 4).(17)

To compare among these bounds,let us take q 2=12

,q 4=0.01,q 1=(1?q 2?q 4)sin 2θ,q 3=(1?q 2?q 4)cos 2

θ,θ∈[0,π4

].From Fig.1we see that our new bound (14)detects entanglement for θ>0.143.While C P P T and C r detect entanglement for θ>0.390and θ>0.613,respectively,C W cannot detect any entanglement as it is always negative.

In summary,by using a positive map we have presented a new lower bound of concurrence for arbitrary dimensional

bipartite systems.By detailed example we have shown that this bound is better for some states than the lower bounds from the PPT criterion,the realignment criterion,and the Breuer’s entanglement witness[12]in detecting quantum entanglement.

ACKNOWLEDGMENTS

This work was supported by the NSFC (10875081,10801100,10871228),KZ200810028013and PHR201007107.

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