Common use of Examples I Clause in Contracts

Examples I. The following examples illustrate the correspondence established in Section 3.1. They show in particular that very often (Examples 1, 2, and 3), but not always (Example 4), the direct connection between entanglement and positive intrinsic information holds with respect to the standard bases (i.e., the bases physicists use by commodity and intuition). Example 1 was already analyzed in [15]. The examples of this section will be discussed further in Section 3.5 under the aspect of the existence of key-agreement protocols in the classical and quantum regimes. Example 1. Let us consider the so-called 4-state protocol of [3]. The analysis of the 6-state protocol [1] is analogous and leads to similar results. We compare the is [11] √ √ √ √ Ψ = F/2 |0, 0)⊗ ξ00 + D/2 |0, 1)⊗ ξ01 + D/2 |1, 0)⊗ ξ10 + F/2 |1, 1)⊗ ξ11 , where D (the disturbance)is the probability that X ƒ= Y holds if X and Y are the classical random variables of ▇▇▇▇▇ and ▇▇▇, respectively, where F = 1 − D (the fidelity ), and where the ξij satisfy (ξ00|ξ11) = (ξ01|ξ10) = 1− 2D and (ξii|ξij) = 0 for all i ƒ= j. Then the state ρAB is (in the basis {| 00 ), | 01 ), | 10 ), | 11 )}) 1  D 0 0 −D(1 − 2D)  ρAB =  0 1 − D −(1 − D)(1 − 2D) 0  2  0 −(1 − D)(1 − 2D) 1 − D 0  −D(1 − 2D) 0 0 D and its partial transpose 1  D 0 0 −(1 − D)(1 − 2D)  = AB 2  0 1 − D −D(1 − 2D) 0  0 −D(1 − 2D) 1 − D 0  −(1 − D)(1 − 2D) 0 0 D has the eigenvalues (1/2)(D (1 D)(1 2D)) and (1/2)((1 D) D(1 2D)), which are all non-negative (i.e., ρAB is separable) if D ≥ 1 − √2 . (4) From the classical viewpoint, the corresponding distributions (arising from measuring the above quantum system in the standard bases) are as follows. First, X and Y are both symmetric bits with Prob [X = Y ] = D. Eve’s random variable Z = [Z1, Z2] is composed of 2 bits Z1 and Z2, where Z1 = X Y , i.e., Z1 tells Eve whether ▇▇▇ received the qubit disturbed (Z1 = 1) or not (Z1 = 0) (this is a consequence of the fact that the ξii and ξij (i =ƒ j)states generate orthogonal sub- spaces), and where the probability that Eve’s seco√nd bit indicates the correct √value of Bob’s bit is Prob[Z2 = Y ] = δ = (1 + 1 − (ξ00|ξ11)2)/2 = 1/2 +

Examples I. The following examples illustrate the correspondence established in Section 3.1. They show in particular that very often (Examples 1, 2, and 3), but not always (Example 4), the direct connection between entanglement and positive intrinsic information holds with respect to the standard bases (i.e., the bases physicists use by commodity and intuition). Example 1 was already analyzed in [15]. The examples of this section will be discussed further in Section 3.5 under the aspect of the existence of key-agreement protocols in the classical and quantum regimes. Example 1. Let us consider the so-called 4-state protocol of [3]. The analysis of the 6-state protocol [1] is analogous and leads to similar results. We compare the is [11] √ √ √ √ Ψ = F/2 |0, 0)⊗ ξ00 + D/2 |0, 1)⊗ ξ01 + D/2 |1, 0)⊗ ξ10 + F/2 |1, 1)⊗ ξ11 , where D (the disturbance)is the probability that X ƒ= =ƒ Y holds if X and Y are the classical random variables of ▇▇▇▇▇ and ▇▇▇, respectively, where F = 1 − D (the fidelity ), and where the ξij satisfy (ξ00|ξ11) = (ξ01|ξ10) = 1− 2D and (ξii|ξij) = 0 for all i ƒ= j. Then the state ρAB is (in the basis {| 00 ), | 01 ), | 10 ), | 11 )}) 1  D 0 0 −D(1 − 2D)  ρAB =  0 1 − D −(1 − D)(1 − 2D) 0  2  0 −(1 − D)(1 − 2D) 1 − D 0  −D(1 − 2D) 0 0 D and its partial transpose 1  D 0 0 −(1 − D)(1 − 2D)  = AB 2  0 1 − D −D(1 − 2D) 0  0 −D(1 − 2D) 1 − D 0  −(1 − D)(1 − 2D) 0 0 D has the eigenvalues (1/2)(D (1 D)(1 2D)) and (1/2)((1 D) D(1 2D)), which are all non-negative (i.e., ρAB is separable) if D ≥ 1 − √2 . (4) From the classical viewpoint, the corresponding distributions (arising from measuring the above quantum system in the standard bases) are as follows. First, X and Y are both symmetric bits with Prob [X = Y ] = D. Eve’s random variable Z = [Z1, Z2] is composed of 2 bits Z1 and Z2, where Z1 = X Y , i.e., Z1 tells Eve whether ▇▇▇ received the qubit disturbed (Z1 = 1) or not (Z1 = 0) (this is a consequence of the fact that the ξii and ξij (i =ƒ ƒ= j)states generate orthogonal sub- spaces), and where the probability that Eve’s seco√nd bit indicates the correct √value of Bob’s bit is Prob[Z2 = Y ] = δ = (1 + 1 − (ξ00|ξ11)2)/2 = 1/2 +