Researching | Loss-tolerant measurement device independent quantum key distribution with reference frame misalignment

Journals >Chinese Optics Letters >Volume 20 >Issue 9 >Page 092701 > Article

Chinese Optics Letters
Vol. 20, Issue 9, 092701 (2022)

Loss-tolerant measurement device independent quantum key distribution with reference frame misalignment

Jipeng Wang, Zhenhua Li, Zhongqi Sun, Tianqi Dou, Wenxiu Qu, Fen Zhou, Yanxin Han, Yuqing Huang, and Haiqiang Ma^*

Author Affiliations

School of Science and State Key Laboratory of Information Photonics and Optical Communications, Beijing University of Posts and Telecommunications, Beijing 100876, China

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DOI: 10.3788/COL202220.092701 Cite this Article Set citation alerts

Jipeng Wang, Zhenhua Li, Zhongqi Sun, Tianqi Dou, Wenxiu Qu, Fen Zhou, Yanxin Han, Yuqing Huang, Haiqiang Ma. Loss-tolerant measurement device independent quantum key distribution with reference frame misalignment[J]. Chinese Optics Letters, 2022, 20(9): 092701 Copy Citation Text

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Abstract

Reference frame independent and measurement device independent quantum key distribution (RFI-MDI-QKD) has the advantages of being immune to detector side loopholes and misalignment of the reference frame. However, several former related research works are based on the unrealistic assumption of perfect source preparation. In this paper, we merge a loss-tolerant method into RFI-MDI-QKD to consider source flaws into key rate estimation and compare it with quantum coin method. Based on a reliable experimental scheme, the joint influence of both source flaws and reference frame misalignment is discussed with consideration of the finite-key effect. The results show that the loss-tolerant RFI-MDI-QKD protocol can reach longer key rate performance while considering the existence of source flaws in a real-world implementation.

Keywords

measurement device independence quantum key distribution source flaw

Q_{i α, j β}^{M_{a} M_{b}, U} = Q_{i α, j β}^{M_{a} M_{b}} (1 + ξ^{l r}), Q_{i α, j β}^{M_{a} M_{b}, L} = Q_{i α, j β}^{M_{a} M_{b}} (1 - ξ^{l r}), - \frac{f (ϵ^{1.5})}{\sqrt{N^{M_{a} M_{b}} Q_{i α, j β}^{M_{a} M_{b}}}} \leq ξ^{M_{a} M_{b}} \leq \frac{f (ϵ^{4} / 16)}{\sqrt{N^{M_{a} M_{b}} Q_{i α, j β}^{M_{a} M_{b}}}} .

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Y_{i α, j β}^{L} = \frac{μ_{a}^{2} μ_{b} S_{1} - v_{a}^{2} v_{b} S_{2}}{μ_{a} μ_{b} v_{a} v_{b} (μ_{a} - v_{a})}, S_{1} = Q_{i α, j β}^{v_{a} ν_{b}, L} e^{(ν_{a} + ν_{b})} + Q_{i α, j β}^{0_{a} 0_{b}, L} - Q_{i α, j β}^{v_{a} 0_{b}, U} e^{v_{a}} - Q_{i α, j β}^{0_{a} v_{b}, U} e^{v_{b}}, S_{2} = Q_{i α, j β}^{μ_{a} μ_{b}, U} e^{(μ_{a} + μ_{b})} + Q_{i α, j β}^{0_{a} 0_{b}, U} - Q_{i α, j β}^{μ_{a} 0_{b}, L} e^{μ_{a}} - Q_{i α, j β}^{0_{a} μ_{b}, L} e^{μ_{b}}, Y_{i α, j β}^{U} = \frac{e^{(v_{a} + v_{b})} Q_{i α, j β}^{0_{a} 0_{b}, U} + Q_{i α, j β}^{v_{a} v_{b}, U} - e^{v_{a}} Q_{i α, j β}^{v_{a} 0_{b}, L} - e^{v_{b}} Q_{i α, j β}^{0_{a} b_{b}, L}}{ν_{a} v_{b}} .

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R_{L} = μ_{a} μ_{b} e^{- (μ_{a} + μ_{b})} (1 - I_{E}) \sum_{α = β = Z} Y_{i α, j β}^{L} - λ_{E C}, I_{E} = (1 - E_{Z Z}^{11, U}) h (\frac{1 + v_{\max}}{2}) + E_{Z Z}^{11, U} h [\frac{1 + f (v_{\max})}{2}], f (v_{\max}) = \frac{\sqrt{\frac{C^{L}}{2} - {(1 - E_{Z Z}^{11, U})}^{2} v_{\max}^{2}}}{E_{Z Z}^{11, U}}, v_{\max} = \min (\frac{1}{1 - E_{Z Z}^{11, U}} \sqrt{\frac{C^{L}}{2}}, 1) .

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C^{L} = \sum_{α, β \in {X, Y}} min: {(1 - 2 E_{α β}^{11})}^{2}, s.t. E_{α β}^{11, L} \leq E_{α β}^{11} \leq E_{α β}^{11, U} .

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| Ψ_{A} 〉 = \frac{1}{\sqrt{2}} ({| 0_{Z} 〉}_{A 1} \otimes {| ϕ_{0 Z} 〉}_{A 2 E} + {| 1_{Z} 〉}_{A 1} \otimes {| ϕ_{1 Z} 〉}_{A 2 E}),

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| Ψ_{A} 〉 = \frac{1}{\sqrt{2}} ({| 0_{X} 〉}_{A 1} \otimes {| ϕ_{0 X}^{vir} 〉}_{A 2 E} + {| 1_{X} 〉}_{A 1} \otimes {| ϕ_{1 X}^{vir} 〉}_{A 2 E}), | Ψ_{A} 〉 = \frac{1}{\sqrt{2}} ({| 0_{Y} 〉}_{A 1} \otimes {| ϕ_{0 Y}^{vir} 〉}_{A 2 E} + {| 1_{Y} 〉}_{A 1} \otimes {| ϕ_{1 Y}^{vir} 〉}_{A 2 E}) .

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Y_{i α, j β}^{vir} = P_{i α}^{vir} P_{j β}^{vir} [S_{σ_{t}^{A}}^{i α, vir} {(S_{σ_{t}^{B}}^{j β, vir})}^{T}] \cdot q_{σ_{t}^{A} \otimes σ_{t}^{B}},

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S_{σ_{t}^{A}}^{i α, vir} = [S_{I^{A}}^{i α, vir}, S_{X^{A}}^{i α, vir}, S_{Y^{A}}^{i α, vir}, S_{Z^{A}}^{i α, vir}], S_{σ_{t}^{B}}^{j β, vir} = [S_{I^{B}}^{j β, vir}, S_{X^{B}}^{j β, vir}, S_{Y^{B}}^{j β, vir}, S_{Z^{B}}^{j β, vir}], q_{σ_{t}^{A} \otimes σ_{t}^{B}} = [q_{I^{A} \otimes I^{B}}, q_{I^{A} \otimes X^{B}}, q_{I^{A} \otimes Y^{B}}, q_{I^{A} \otimes Z^{B}}, q_{X^{A} \otimes I^{B}}, q_{X^{A} \otimes X^{B}}, q_{X^{A} \otimes Y^{B}}, q_{X^{A} \otimes Z^{B}}, q_{Y^{A} \otimes I^{B}}, q_{Y^{A} \otimes X^{B}}, q_{Y^{A} \otimes Y^{B}}, q_{Y^{A} \otimes Z^{B}}, q_{Z^{A} \otimes I^{B}}, q_{Z^{A} \otimes X^{B}}, q_{Z^{A} \otimes Y^{B}}, q_{Z^{A} \otimes Z^{B}}] .

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q_{σ_{t}^{A} \otimes σ_{t}^{B}} = \frac{Y_{i α, j β}}{P_{i α} P_{j β} [S_{σ_{t}^{A}}^{i α} {(S_{σ_{t}^{B}}^{j β})}^{T}]} .

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{| ϕ_{0 Z} 〉}_{A} = | 0_{Z} 〉, {| ϕ_{1 Z} 〉}_{A} = | 1_{Z} 〉, {| ϕ_{0 X} 〉}_{A} = \frac{| 0_{Z} 〉 + | 1_{Z} 〉}{\sqrt{2}} = | 0_{X} 〉, {| ϕ_{0 Y} 〉}_{A} = \frac{| 0_{Z} 〉 + i | 1_{Z} 〉}{\sqrt{2}} = | 0_{Y} 〉,

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{| ϕ_{0 Z} 〉}_{B} = \cos (δ_{1}) | 0_{Z} 〉 + \sin (δ_{1}) | 1_{Z} 〉, {| ϕ_{1 Z} 〉}_{B} = \sin (δ_{2}) | 0_{Z} 〉 + \cos (δ_{2}) | 1_{Z} 〉, {| ϕ_{0 X} 〉}_{B} = \sin (\frac{π}{4} + δ_{3}) | 0_{Z} 〉 + \cos (\frac{π}{4} + δ_{3}) e^{i (θ_{1} + ω)} | 1_{Z} 〉, {| ϕ_{0 Y} 〉}_{B} = \sin (\frac{π}{4} + δ_{3}) | 0_{Z} 〉 + \cos (\frac{π}{4} + δ_{3}) e^{i (\frac{π}{2} + θ_{2} + ω)} | 1_{Z} 〉 .

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Q_{0} = \frac{1}{2 π} \int_{0}^{2 π} 1 - (1 - e_{dark}) \exp (- {| A + e^{i Δ ϕ} B |}^{2}) d Δ ϕ, Q_{1} = \frac{1}{2 π} \int_{0}^{2 π} 1 - (1 - e_{dark}) \exp (- {| A + e^{i π} e^{i Δ ϕ} B |}^{2}) d Δ ϕ,

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Jipeng Wang, Zhenhua Li, Zhongqi Sun, Tianqi Dou, Wenxiu Qu, Fen Zhou, Yanxin Han, Yuqing Huang, Haiqiang Ma. Loss-tolerant measurement device independent quantum key distribution with reference frame misalignment[J]. Chinese Optics Letters, 2022, 20(9): 092701

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