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Bell’s theorem without inequalities for one class of two-qubit mixed states Lin-mei Liang a,b,∗ , Cheng-zu Li a a Department of Applied Physics, National University of Defense Technology, Changsha 410073, PR China b Laboratory of Quantum Communication and Quantum Computation, University of Science and Technology of China,

Hefei 230026, PR China Received 5 May 2003; received in revised form 9 September 2003; accepted 15 September 2003 Communicated by P.R. Holland

Abstract This Letter presents a new kind of proof to show Bell’s theorem without inequalities for one class of mixed entangled states of two qubits. 2003 Elsevier B.V. All rights reserved. PACS: 03.65.Ud; 03.65.Ta; 03.67.-a; 42.50.-p

Bell’s theorem [1] states that one cannot in general reproduce the results of quantum theory with a classical, deterministic local model, i.e., refutes local theories based on Einstein, Podolsky, and Rosen’s “elements of reality” [2]. Greenberger, Horne, and Zeilinger (GHZ) [3] caused much interest when they gave a proof of non-locality but without using inequalities. Their proof requires three or more particles. Hardy’s argument of ‘non-locality without inequalities’ for two particles is considered ‘the best version of Bell’s theorem’ which is hold for nonmaximally pure entangled states [4]. Based on Hardy’s work, Goldstein proposes a more concise Hardy theorem [5]. Then Cabello presents Bell’s theorem without inequalities for two observers which is hold for

* Corresponding author.

E-mail address: [email protected] (L.-m. Liang). 0375-9601/$ – see front matter 2003 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2003.09.038

maximally entangled states [6,7]. However, pure entangled states, in general, is an ideal state. In actual environment, there are noise everywhere, pure entangled states unavoidably becomes mixed states because of decoherence with noise. Thus it is also important to show Bell’s theorem without inequalities for mixed states. Kar shows that mixed states of three or more qubits permit a Hardy-like proof without inequalities of Bell’s theorem [8,9]. In this Letter, we will present a new kind of proof to show Bell’s theorem without inequalities for one class of mixed entangled states of two qubits. Consider a two-qubit mixed state ρ = p|Ψ Ψ | + (1 − p)|1111| (p = 0),

(1)

where |Ψ is Hardy state [5], i.e., |Ψ = a|00 + b|10 + c|01 and abc = 0, |0, |1 is orthonormal bases of qubit 1 and qubit 2, and p is the probability of Hardy state in the mixed state. In the following, we

L.-m. Liang, C.-z. Li / Physics Letters A 318 (2003) 300–302 +|ab| +|bc| will show that if p > |abc||ac| 2 +|ac|2 +|ab|2 +|bc|2 , the state ρ can present nonlocality without inequalities. Let us first define the following Hermite operator: 2

2

2

Ui = |11|,

(2a)

Wi = |ββ|,

(2b)

Vi = |γ γ |,

(2c)

where a|0 + b|1 , |β = |a|2 + |b|2 a|0 + c|1 |γ = , |a|2 + |c|2

(3a)

and U1 = 0|ρ|U1 = 0 = p(|a|2 + |c|2 )|V2 = 1 × V2 = 1|, then the condition that abc = 0 and p = 0 guarantee that if U1 = 0, then one can get V2 = 1. Likewise, Pρ (W1 = 1|U2 = 0) = 1 can be obtained. For Pρ (W1 = V2 = 0), it is easy to calculate that W1 = V2 = 0|ρ|W1 = V2 = 0 = p1 . However, according to EPR, the probability of obtaining U1 U2 = 0 is p, and if U1 = 0, the probability of obtaining V2 = 1 is 1, and the probability of W1 = 1 under the condition that U2 = 0 is 1, then according to local realistic theory that Pρ (W1 = 1 or V2 = 1) p, i.e., Pρ (W1 = V2 = 0) should be equal to or smaller +|ab| +|bc| than 1 − p. But, if p > |abc||ac| 2 +|ac|2 +|ab|2 +|bc|2 = p0 , quantum mechanics gives that Pρ (W1 = V2 = 0) = p1 > 1 − p, which contradicts local realistic theory. Therefore property (4a)–(4d) cannot be explained by local realistic theory, i.e., if p > p0 , the state ρ defined in Eq. (1) can present non-locality. Because the state ρ defined in Eq. (1) under the condition that p > p0 presents non-locality without inequalities, the state ρ must be entangled. However, we cannot conclude that when p p0 , the state ρ is separable. In fact, according to the separability condition obtained by Peres [10] and Horodecki [11], if the state ρ is separable, then all the eigenvalues of the partial transposition of ρ, ρ PT 2

(3b)

and Ui , Wi , Vi refers the operator on qubit i (i = 1, 2). For the state ρ defined in Eq. (1), one can easily find that Pρ (U1 U2 = 0) = p,

(4a)

Pρ (V2 = 1|U1 = 0) = 1,

(4b)

Pρ (W1 = 1|U2 = 0) = 1, Pρ (W1 = V2 = 0) = p1 =

301

(4c) p|abc|2 + (1 − p)|a|4 (|a|2 + |b|2)(|a|2 + |c|2)

,

(4d) where Pρ (U1 U2 = 0) means the probability of obtaining the result 0 when measuring U1 on qubit 1 or measuring U2 on qubit 2, and Pρ (V2 = 1|U1 = 0) = 1 means that if one measures U1 on qubit 1 and obtains the result 0, then qubit 2 will give the result 1 with certainty when V2 is measured. Likewise, Pρ (W1 = 1|U2 = 0) = 1 means that when one gets the result 0 on qubit 2 by measuring U2 , then the qubit 1 will definitely shows the result 1 when W1 is measured. And Eq. (4d) points out that when we measure W1 on qubit 1 and V2 on qubit 2, then the probability that we can simultaneously get the result 0 on these two qubits is p1 . Pρ (U1 U2 = 0) = p can be obtained directly, because there is no |11 term in the state |Ψ . Eq. (4b) can be obtained as follows: ρ can also be written as ρ = p |a|2 + |c|2 |0γ + b|10 × |a|2 + |c|2 0γ | + b∗ 10| + (1 − p)|1111| (p = 0)

(5)

p|a|2 pac∗ PT ρ = ∗ pba ∗ pbc

pca ∗ p|c|2 0 0

pab∗ 0 p|b|2 0

2

pcb∗ 0 0 1−p

2

(6)

are non-negative. In other words, the state ρ is separable if and only if that all principal minors of the partial transposition of ρ should be non-negative. The above two ways of expression are equivalent which is guaranteed by Sylvester’s criterion [12]. For the state ρ defined in Eq. (1), it is separable if and only if the following four inequalities should hold true: p|a|2 0, p|a|2 pca ∗ pac∗ p|c|2 0, p|a|2 pca ∗ pab∗ pac∗ p|c|2 0 = −p3 |abc|2 0, pba ∗ 0 p|b|2

(7a) (7b)

(7c)

302

L.-m. Liang, C.-z. Li / Physics Letters A 318 (2003) 300–302

p|a|2 pca ∗ pab∗ pcb∗ pac∗ p|c|2 0 0 0. (7d) pba ∗ 0 p|b|2 0 0 0 1−p pbc∗ It is easy to find that the above four inequalities hold true if and only if p = 0. So for the state ρ defined in Eq. (1), if 0 < p 1, the state ρ is entangled. Therefore, if a state ρ is entangled, it is not necessarily that ρ can also present non-locality without inequalities in some sense. But if ρ presents non-locality without inequalities, the state ρ should be entangled. In summary, we present a new kind of proof to show Bell’s theorem without inequalities for one class of mixed entangled states of two qubits. The proof features a novel logical structure. It is not a Hardylike proof [4] based on the logical structure: A and B sometimes happen, A always implies D, B always implies C, but C and D never happen. It is not a Cabello-like proof [13] based on the logical structure: A and B sometimes happen, A always implies D, B always implies C, but C and D happen with a lower probability than A and B. The logical structure of the proof provided in this Letter is: A or B sometimes happen, A always implies D, B always implies C, but neither C nor D happen with a higher probability than one minus the probability of A or B.

Acknowledgements We are very grateful to the referee for his useful suggestion.

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