Common use of A Classical Measure for Quantum Entanglement Clause in Contracts

A Classical Measure for Quantum Entanglement. It is a challenging problem of theoretical quantum physics to find good measures for entanglement [26]. Corollary 3 above suggests the following measure, which is based on classical information theory. Definition 1 Let for a quantum state ρAB µ(ρAB):= min ( max (I(X; Y↓Z))) , {|z⟩} {|x⟩},{|y⟩} where the minimum is taken over all Ψ = Σ √p ψ ⊗ |z⟩ such that ρ = z z AB TrHE (PΨ ) holds and over all generating sets {|z⟩} of HE, the maximum is over all bases {|x⟩} of HA and {|y⟩} of HB, and where PXY Z(x, y, z):= |⟨x, y, z|Ψ⟩|2. ❢ The function µ has all the properties required from such a measure. If ρAB is pure, i.e., ρAB = |ψΣAB ⟩⟨ψAB|, then we have in the Xxxxxxx basis (see for example [24]) ψAB = j cj|xj, yj⟩, and µ(ρAB) = −Tr(ρA log ρA) (where ρA = TrB(ρAB)) as it should [26]. It is obvious that µ is convex, i.e., µ(λρ1 + (1 − λ)ρ2) ≤ λµ(ρ1)+ (1 − λ)µ(ρ2). (1 (−) √λ)/4 001 + 012 + 103 + 114 , where ψ = 10 01 / 2, and ρ = E√xample 5. This example is based on Xxxxxx’x states. Let Ψ = √λψ(−) ⊗ |0⟩ + − | ⟩ | − ⟩ AB λPψ(−) + ((1 − λ)/4)11. It is well-known that ρAB is separable if and only if λ ≤ 1/3. Then the classical distribution is P (010)= P (100)= λ/2 and P (001)= P (012)= P (103)= P (114)= (1 − λ)/4. If λ ≤ 1/3, then consider the channel PZ|Z (0, 0)= PZ|Z (2, 2)= PZ|Z (3, 3)= 1 , PZ|Z (0, 1) = PZ|Z (0, 4) = ξ, PZ|Z (1, 1) = PZ|Z (4, 4) = 1 − ξ, where ξ = 2λ/(1 −λ) ≤ 1. Then µ(ρAB)= I(X; Y↓Z)= I(X; Y |Z)= 0 holds, as it should. If λ > 1/3, then consider the (obviously optimal) channel PZ|Z (0, 0) = PZ|Z (2, 2)= PZ|Z (3, 3)= PZ|Z (0, 1)= PZ|Z (0, 4)= 1. Then µ(ρAB)= I(X; Y↓Z)= I(X; Y |Z)= PZ (0) · I(X; Y |Z = 0)‌ = 1+ λ · (1 − q log2 q − (1 − q)log2(1 − q)) , where q = 2λ/(1 + λ). ♦

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Samples: Classical and Quantum Key Agreement, Classical and Quantum Key Agreement

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A Classical Measure for Quantum Entanglement. It is a challenging problem of theoretical quantum physics to find good measures for entanglement [26]. Corollary 3 above suggests the following measure, which is based on classical information theory. Definition 1 Let for a quantum state ρAB µ(ρAB):= µ(ρAB) := min ( max (I(X; Y↓ZY ↓Z))) , {|z⟩|z)} {|x⟩},{|y⟩|x)},{|y)} Σ where the minimum is taken over all Ψ = Σ √p ψ z √pzψz |z⟩ |z) such that ρ ρAB = z z AB TrHE (PΨ ) holds and over all generating sets {|z⟩|z)} of HE, the maximum is over all bases {|x⟩|x)} of HA and {|y⟩|y)} of HB, and where PXY Z(x, y, z):= |⟨xz) := |(x, y, z|Ψ⟩|2z|Ψ )|2. ρAB is pure, i.e., ρAB = |ψΣAB )(ψAB|, then we have in the Xxxxxxx basis (see The function µ has all the properties required from such a measure. If ρAB is pure, i.e., ρAB = |ψΣAB ⟩⟨ψAB|, then we have in the Xxxxxxx basis (see for example [24]) ψAB = j cj|xj, yj⟩yj), and µ(ρAB) = −Tr(ρA log ρA) (where ρA = TrB(ρAB)) as it should [26]. It is obvious that µ is convex, i.e., µ(λρ1 + (1 − λ)ρ2) ≤ λµ(ρ1)+ λµ(ρ1) + (1 − λ)µ(ρ2). (1 (−) √λ)/4 001 + 012 + 103 + 114 , where ψ = 10 01 / 2, and ρ = E√xample 5. This example is based on Xxxxxx’x states. Let Ψ = √λψ(−λψ − ⊗ |0) ⊗ |0⟩ + (1 | ⟩ | λ)/4 |001 + 012 + 103 + 114), where ψ(−) = |10 ⟩ AB 01)/√2, and ρAB = λPψ(−) + ((1 − λ)/4)11. It is well-known that ρAB is separable if and only if λ ≤ 1/3. Then the classical distribution is P (010)= 010) = P (100)= 100) = λ/2 and P (001)= 001) = P (012)= 012) = P (103)= 103) = P (114)= 114) = (1 − λ)/4. If λ ≤ 1/3, then consider the channel PZ|Z (0, 0)= 0) = PZ|Z (2, 2)= 2) = PZ|Z (3, 3)= 3) = 1 , PZ|Z (0, 1) = PZ|Z (0, 4) = ξξ , PZ|Z (1, 1) = PZ|Z (4, 4) = 1 − ξξ , where ξ = 2λ/(1 −λ− λ) ≤ 1. Then µ(ρAB)= I(X; Y↓Z)= µ(ρAB) = I(X; Y |Z)= ↓Z) = I(X; Y |Z) = 0 holds, as it should. If λ > 1/3, then consider the (obviously optimal) channel PZ|Z (0, 0) = PZ|Z (2, 2)= 2) = PZ|Z (3, 3)= 3) = PZ|Z (0, 1)= 1) = PZ|Z (0, 4)= 4) = 1. Then µ(ρAB)= I(X; Y↓Z)= µ(ρAB) = I(X; Y |Z)= ↓Z) = I(X; Y |Z) = PZ (0) · I(X; Y |Z = 0)‌ 0) = 11 + λ · (1 − q log2 log q − (1 − q)log2(1 q) log (1 − q)) , where q = 2λ/(1 + λ). ♦

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Samples: Classical and Quantum Key Agreement

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A Classical Measure for Quantum Entanglement. It is a challenging problem of theoretical quantum physics to find good measures for entanglement [26]. Corollary 3 above suggests the following measure, which is based on classical information theory. Definition 1 Let for a quantum state ρAB µ(ρAB):= µ(ρAB) := min ( max (I(X; Y↓Z))) , {|z⟩|z)} {|x⟩},{|y⟩|x)},{|y)} z z z AB where the minimum is taken over all Ψ = Σ √p ψ ⊗ |z⟩ |z) such that ρ = z z AB TrHE () holds and over all generating sets bases {|z⟩|z)} of HE, the maximum is over all bases {|x⟩|x)} of HA and {|y⟩|y)} of HB, and where PXY Z(x, y, z):= |⟨xz) := |(x, y, z|Ψ⟩|2z|Ψ)|2 . ❢ The function g Then µ has all the properties required from such a measure. If ρAB is pure, i.e., ρAB = |ψΣAB ⟩⟨ψAB||ψABΣ)(ψAB|, then we have in the Xxxxxxx ScΣhmidt basis (see for example [24]) ψAB = j cj|xj, yj⟩yj), and µ(ρAB) = −Tr(ρA log ρA− j |cj|2 log2(|cj|2) (where ρA = TrB(ρAB−Tr(ρAB log2 ρAB)) , as it should [26]. It is obvious that µ is convex, i.e., µ(λρ1 + (1 − λ)ρ2) ≤ λµ(ρ1)+ (1 − λ)µ(ρ2). Example 4 (1 (−) √λ)/4 001 + 012 + 103 + 114 , where ψ = 10 01 / 2, and ρ = E√xample 5. This example is based on Xxxxxx’x states). Let Ψ = √λψ(−√λψ(−)⊗|0) ⊗ |0⟩ √ − λ)/4 |001+ − | ⟩ | − ⟩ AB √ + (1 012+103+114), where ψ(−) = |10−01)/ 2, and ρAB = λPψ(−) + +((1 − λ)/4)111−λ)/4)11. It is well-known that ρAB is separable if and only if λ 1/3. Then the classical clas- sical distribution is P (010)= P 010) = (100)= 100) = λ/2 and P (001)= P 001) = (012)= P 012) = (103)= P 103) = (114)= 114) = (1 − λ)/4. If λ ≤ 1/3, then consider the channel PZ|Z (0, 0)= 0) = PZ|Z (2, 2)= 2) = PZ|Z (3, 3)= 3) = 1 , PZ|Z (0, 1) = PZ|Z (0, 4) = ξ, PZ|Z (1, 1) = PZ|Z (4, 4) = 1 − ξ, where − ≤ ↓ | ξ = 2λ/(1 −λλ) 1. Then µ(ρAB)= I(X; Y↓Z)= µ(ρAB) = I(X; Y |Z)= Z) = I(X; Y Z) = 0 holds, as it should. If λ > 1/3, then consider the (obviously optimal) channel PZ|Z (0, 0) = PZ|Z (2, 2)= 2) = PZ|Z (3, 3)= 3) = PZ|Z (0, 1)= 1) = PZ|Z (0, 4)= 4) = 1. Then µ(ρAB)= 1+ λ µ(ρAB) = I(X; Y↓Z)= Y↓Z) = I(X; Y |Z)= |Z) = PZ (0) · I(X0)·I(X; Y |Z = 0)‌ 0) = 1+ λ · 2 ·(1 − q 1−q log2 q − (1 − q)log2(1 − qq−(1−q) log2(1−q)) , where q = 2λ/(1 + λ). ♦

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Samples: Classical and Quantum Key Agreement

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