Comment on: “A class of bound entangled states” [Phys. Lett. A 352 (2006) 321]

Comment on: “A class of bound entangled states” [Phys. Lett. A 352 (2006) 321]

Physics Letters A 364 (2007) 517–521 www.elsevier.com/locate/pla Comment Comment on: “A class of bound entangled states” [Phys. Lett. A 352 (2006) 3...

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Physics Letters A 364 (2007) 517–521 www.elsevier.com/locate/pla

Comment

Comment on: “A class of bound entangled states” [Phys. Lett. A 352 (2006) 321] Wei Cheng College of Computer Science and Engineering, University of Electronic Science and Technology of China, Chengdu 610054, China Received 11 October 2006; received in revised form 6 December 2006; accepted 28 December 2006 Available online 18 January 2007 Communicated by P.R. Holland

Abstract We generalize the condition keeping the special states bound entangled [S.-M. Fei, X. Li-Jost, B.-Z. Sun, Phys. Lett. A 352 (2006) 321] to the general states. Taking the original 4 × 4 states as 2 × 8 states we show that they still can be bound entangled. © 2007 Elsevier B.V. All rights reserved. PACS: 03.65.Bz; 89.70.+c Keywords: Bound entangled states; PPT

1. Introduction

Construction of PPTES [1,2] is a non-trivial task, and the UPB construction is really the only known automatic procedure [3]. Recently, in Ref. [4], S.-M. Fei et al. have constructed a class of bound entangled states. In this comment, relating on the construction of the 4 × 4 states (4) in Ref. [4], we first generalize the condition keeping the special states (7) in Ref. [4] bound entangled to the general states (4) in Ref. [4], then we take the 4 × 4 states (4) in Ref. [4] as 2 × 8 states and show that they still violate the range criterion with a different parameter range of ε from that of S.-M. Fei et al. in Ref. [4], thus present a new class of bound entangled states.

2. The condition 0  ε 

1 2

is universal

We first recall the result in Ref. [4]. S.-M. Fei et al. consider the following states, i.e., the states (4) in Ref. [4]:

DOI of original article: 10.1016/j.physleta.2005.12.038. E-mail address: [email protected]. 0375-9601/$ – see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2006.12.077

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x1 ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ρ=⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎣ 0 0

0 x3 0 0 −x3 0 0 0 0 0 0 0 0 0 0 0

0 0 x2 0 0 0 0 0 −x2 0 0 0 0 0 0 0

0 0 0 −x3 0 0 0 0 0 x3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 x1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 −x2 0 0 0 0 0 0 0 0 0 0 0 0 0 x5 0 0 0 x2 0 0 0 0 0 0 0 0 0 0 0 0 0 −x5 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 x1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 x4 0 0 −x4 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 −x5 0 0 0 0 0 x5 0 0

0 0 0 0 0 0 0 0 0 0 0 −x4 0 0 x4 0

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥, 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎦ x1

ε ε ε ε 2 2 2 2 where x1 = 1−ε 4 , x2 = 2 |a| , x3 = 2 |b| , x4 = 2 |c| , x5 = 2 |d| . First, there is a typing error in the inequality (6) in Ref. [4]. It should be “”, not “>”. Then it should be  1 − ε ε 4 2 |a| + |b|4 + |c|4 + |d|4 + Δ1  0, − 4 4

(1)

(2)

where Δ1 = [(|a|2 − |d|2 )2 + (|b|2 + |c|2 )2 ][(|a|2 + |d|2 )2 + (|b|2 − |c|2 )2 ]. Next, since |ψ±a  = [ 0 0 ±a 0 0 0 0 d ∓a 0 0 0 0 −d 0 0 ]T are two pure states [4], then ψ±a ||ψ±a  = 1. So it follows that 1 |a|2 + |d|2 = . 2

(3)

Furthermore, by using |a|2 + |d|2  2|a||d| we can obtain 1 |a||d|  . 4

(4)

Similarly, since |ψ±b  = [ 0 ±b 0 0 ∓b 0 0 0 0 0 0 −c 0 0 c 0 ]T are also two pure states [4], then ψ±b ||ψ±b  = 1. So it follows that 1 |b|2 + |c|2 = . 2

(5)

This together with |b|2 + |c|2  2|b||c| yields 1 |b||c|  . 4 Now we consider the above inequality (2). On one hand, we have 2

2  2 2

2  

|a| + |d|2 + |b|2 − |c|2 Δ1 = |a|2 − |d|2 + |b|2 + |c|2 

2 2  2 2

2 

= |a|2 + |d|2 − 4|a|2 |d|2 + |b|2 + |c|2 |a| + |d|2 + |b|2 + |c|2 − 4|b|2 |c|2 .

(6)

Using (3) and (5), we have Δ1 = [ 12 − 4|a|2 |d|2 ][ 12 − 4|b|2 |c|2 ] = [ 12 − 4(|a||d|)2 ][ 12 − 4(|b||c|)2 ]. According to (4) and (6), we 1 obtain Δ1  14 × 14 = 16 . On the other hand, we have 2

2

|a|4 + |b|4 + |c|4 + |d|4 = |a|2 + |d|2 + |b|2 + |c|2 − 2|a|2 |d|2 − 2|b|2 |c|2 . Using (3) and (5), we have |a|4 + |b|4 + |c|4 + |d|4 = |c|4 + |d|4  12 − 18 − 18 = 14 . So we can get that

1 2

 

 Δ2 = 2 |a|4 + |b|4 + |c|4 + |d|4 + Δ1 

− 2(|a||d|)2 − 2(|b||c|)2 . According to (4) and (6), we obtain |a|4 + |b|4 +





1 2× + 4



1 16

 = 1,

i.e., Δ2  1. Then we can rewrite the inequality (2) as

(7) 1−ε 4



ε 4 Δ2

 0. By using (7), we have ε 

1 2.

W. Cheng / Physics Letters A 364 (2007) 517–521

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So we can conclude that all the roots of (5) in Ref. [4] are positive when 0  ε  12 . Therefore, the states (4) in Ref. [4] are bound entangled when 0  ε  12 . Remark 1. We prove that the state (4) in Ref. [4] is also semi-positive definite under the condition which makes the special state (7) in Ref. [4] semi-positive definite. From the above proof we can also see that the state (4) in Ref. [4] is also a bound entangled state for arbitrary a, b, c and d with this condition. So it is fair to say that the observation presented above provides a complement to S.-M. Fei et al.’s argument. 3. A new class of bound entangled states In this section we first take the 4 × 4 states ρ of (4) in Ref. [4] as 2 × 8 states and find the parameter range of ε under which ρ T2 is positive semidefinite. Then we show that the 2 × 8 states still violate the range criterion. If we take the 4 × 4 states (1) as 2 × 8 states, then we can rewrite the state ρ as ⎤ ⎡ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x1 ⎢ 0 0 0 −x3 0 0 0 0 0 0 0 0 0 0 0 ⎥ x3 ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 −x 0 0 0 0 0 0 0 0 x 2 2 ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ 0 −x3 0 0 x 0 0 0 0 0 0 0 0 0 0 0 3 ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x 1 ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 −x 0 0 0 0 0 0 0 0 x 5 5 ⎥. ⎢ ρ=⎢ (8) ⎥ 0 0 −x 0 0 0 0 0 x 0 0 0 0 0 0 0 2 2 ⎥ ⎢ ⎢ 0 0 0 0 0 0 0 0 0 ⎥ 0 0 0 0 0 0 0 ⎥ ⎢ ⎢ 0 0 0 x 0 0 0 0 0 ⎥ 0 0 0 0 0 0 0 1 ⎥ ⎢ ⎢ 0 0 0 0 x4 0 0 −x4 0 ⎥ 0 0 0 0 0 0 0 ⎥ ⎢ ⎢ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ⎥ ⎥ ⎢ ⎢ 0 0 0 0 0 x5 0 0 ⎥ 0 0 0 0 0 0 −x5 0 ⎥ ⎢ ⎣ 0 0 0 0 −x4 0 0 x4 0 ⎦ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x1 0 0 0 0 0 0 0 0 By the PPT criterion [5,6], we can obtain the eigenvalues of ρ T2 as follows ε 2 ε 2 ε ε 1 − ε − 2ε|a|2 1 − ε + 2ε|a|2 |a| , |a| , ε|b|2 , ε|c|2 , |d|2 , |d|2 , , , 2 2 2 2 4 4 1 − ε − 2ε|d|2 1 − ε + 2ε|d|2 , . 4 4

0, 0, 0, 0, 0, 0,

So we can obtain that ρ T2 is positive semidefinite when   1 1 0  ε  min , . 1 + 2|a|2 1 + 2|d|2 Note that |a|2 + |d|2 = 12 , we have |a|2 

1 2

(10)

and |d|2  12 . This together with (10) yields

1 0ε . 2 If we take a = b = c = d =

(9)

(11) 1 2

in (10), then we can obtain

2 0ε . 3

(12)

Remark 2. Comparing with the calculation result in Ref. [4], we can find that if we take the 4 × 4 states (4) in Ref. [4] as 2 × 8 states, then we can obtain the analytical expression (9) of the eigenvalues of ρ T2 , which is different from that of Ref. [4]. Furthermore, we also reveal the new and concise condition (10) under which the state ρ is bound entangled. Moreover, in the case of the states (7) in Ref. [4], if we take them as 2 × 8 states, then they violate the range criterion when 0  ε  23 . However, as shown by S.-M. Fei et al. in Ref. [4], if we take them as 4 × 4 states, then they violate the range criterion when 0  ε  12 . Finally, let us to show that the 2 × 8 states (8) in this comment violate the range criterion.

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Following a similar strategy as in Ref. [4] we can obtain that the following vectors (1, 0)T ⊗ (A, B, 0, 0, −B, D, 0, 0)T ,   C E E T , (A, −C)T ⊗ 1, 0, , 0, 0, , 0, A C A

(13) and

(14)

(0, 1)T ⊗ (0, 0, F, G, 0, 0, −G, H )T

(15)

span the separable linear independent vectors of the range of ρ, where A, B, C, D, E, F, G, H ∈ Z, and Z stands for complex numbers. All partial complex conjugations of vectors (13), (14), and (15) are (1, 0)T ⊗ (A∗ , B ∗ , 0, 0, −B ∗ , D ∗ , 0, 0)T ,   C∗ E∗ E∗ T (A, −C)T ⊗ 1, 0, ∗ , 0, 0, ∗ , 0, ∗ , A C A

(16) and

(17)

(0, 1)T ⊗ (0, 0, F ∗ , G∗ , 0, 0, −G∗ , H ∗ )T .

(18)

Similarly, we can obtain that the following vectors (1, 0)T ⊗ (A, B, C, 0, −B, D, 0, E)T ,   F T 1, ⊗ (A, 0, C, 0, 0, D, 0, E)T , A

(19) (20)

and

(0, 1)T ⊗ (F, 0, G, H, 0, I, −H, J )T

(21)

span the separable linear independent vectors of the range of ρ T2 , where A, B, C, D, E, F, G, H, I, J ∈ Z, and Z stands for complex numbers. It is now straightforward to check that the vectors (13), (14), and (15) span the separable linear independent vectors of the range T2 of ρ, but the partial complex  of these vectors, i.e., the vectors (16), (17), and (18), do not span the range of ρ . Hence  1 conjugations 1 for any 0  ε  min 1+2|a|2 , 1+2|d|2 , the states (8) in this comment violate the range criterion. Thus the states (8) are bound entangled states. Specially, if we take a = b = c = d = 12 , then for any 0  ε  23 , the states (8) are bound entangled states. Remark 3. In Ref. [4], S.-M. Fei et al. take the states (4) in Ref. [4] as 4 × 4 states and show that they violate the range criterion. Here we take the states (4) in Ref. [4] as 2 × 8 states and also show that they violate the range criterion. Although the 2 × 8 states originate from the 4 × 4 states of (4) in Ref. [4], in fact, they are different essentially. For instance, the 2 × 8 partial transposition states ρ T2 are different from the 4 × 4 partial transposition states ρ T2 . Thus, we give certainly some new bound entangled states. On the other hand, it may be an interesting result that different systems, having the same matrix representation, have similar properties. Remark 4. In the case of a = b = c = d = 12 , it is verified that the trace norm of the realigned matrix of the 2 × 8 states is less than 1 when 0  ε  1. Hence the realignment separability criterion [7,8] could not detect the entanglement of this bound entangled state. On the other hand, we can obtain that the trace norm of partial transposed matrix of the 2 × 8 states is  T  1 ρ 2  = |3ε − 2| + 1 |ε − 2| + ε. 4 4 From the above equality we can also see that the trace norm of the partial transposed matrix ρ T2 of the 2 × 8 bound entangled states is  T  ρ 2  = 1 if 0  ε  2 3

  3 and ρ T2  = ε 2

if

2 < ε  1. 3

Therefore, when 0  ε  23 , neither the lower bound of concurrence nor the lower bound for the entanglement of formation [9,10] could detect the entanglement. Acknowledgements We would like to thank Dr. Shao-Ming Fei, Dr. P.R. Holland and the anonymous referees for their invaluable comments on the manuscript which were very useful for improving the quality of this paper. This work was supported by the Youth Science and Technology Foundation of UESTC and the National Fundamental Research Program of China (No. 2004CB318103).

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