Corrigendum: One Step Quantum Key Distribution Based on EPR Entanglement

Scientific Reports, Apr 2017

Jian Li, Na Li, Lei-Lei Li, Tao Wang

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Corrigendum: One Step Quantum Key Distribution Based on EPR Entanglement

Corrigendum: One Step Quantum Key Distribution Based on EPR Entanglement Jian Li Na Li Lei-Lei Li Tao Wang In the Article, we did not take into account the entanglement swapping that can be caused by eavesdropping efforts. Due to entanglement swapping, in Group 2 of Table 3 and Group 3 of Table 4, the initial states will never lfeoardmtsoaammeaesausurermemenentutsoiuntgcoanmienbcoyrErevcet ?se1+q3ue?n?2c4e,atnhdis w?1?il3l n?e?v24er. lSeiamditloarl?y,1+2if a?ft?3e4r Eanvde's?m1+2eas?u?3r4e m.TehnetrBefoobreptehr-e outcomes for Group 2, Table 3, rows ?Bell states Eve measures and sends? and ?Bell states Bob measures?, and Group 3, Table 4, row ?Bell states Bob measures?, need to be corrected. The correct versions of Table 3 and Table 4 are published below as Table?1 and Table?2 respectively. Number of classical bits Alice?s random bit Alice sending Bell states Eve random measuring basis Eve selects right or wrong location Bell states Eve measures and sends Bob random measuring basis Bell states Bob measures Public discussion of location Public discussion of states Share secret key Errors in key Number of classical bits Alice?s random bit Alice sending Bell states Eve random measuring basis Eve selects right or wrong location Bell states Eve measures and sends Bob random measuring basis Bell states Bob measures Public discussion of location Public discussion of states Share secret key Errors in key Group 1 Group 2 1 2 3 4 1 2 1 0 ?1?3 ?2?4 Bell(13) Bell(24) right ?1?3 ?2?4 ?1?3 ?2?4 ?1+3 ? 2+4 Bell(13) Bell(24) Bell(13) Bell(24) Bell(12) Bell(12) ?1?3 ?2?4 ?1?3 ?2?4 ?1?2 ?1?2 + ?13 Bel(34) ? 3?4 ? 3+4 ? 3+4 Bell(13) Bell(24) 1 ? 1 ? ? ? 1 2 3 4 1 2 ? 3+4 ? 3+4 Bell(13) Bell(24) ? ? right right/wrong ? ? ? ? ? ? 0 ? 3 1 ? ? 3 1 1 ? ? ? right wrong Group 4 0 ? right right 4 0 ? 2+4 ? 3?4 ? ? Furthermore, these changes in the outcomes described in Table 3 and Table 4 lead to different results of the error rate evaluation. If Eve selects the wrong location to measure in the Bell basis, this will lead to entanglement swapping. The initial states ?1+2 ?3+4 will become |?1+2?|?3+4? = 1 (|?1+3?|?2+4? + |??13?|??24? + |?1+3?|?+24? + |?1?3?|??24?) 2 Suppose that if Eve?s measurement yields |?1+3?|?2+4?, |?1+3?|?2+4? = 1 (|?1+2?|?3+4? + |?1?2?|??34? + |?1+2?|?3+4? + |?1?2?|??34?), 2 then Bob performing a measurement using a correct sequence after Eve?s measurement using an incorrect s e q u e n c e w i l l y i e l d o n e o f t h e f o l l o w i n g f o u r p o s s i b l e r e s u l t s w i t h e q u a l p r o b a b i l i t y : ?1+2 ?3+4 , ?1?2 ??34 , ?1+2 ?3+4 , ?1?2 ??34 . Similarly, we can expand the other three possible outcomes of Eve?s measurement to |?1?3?|??24? = 1 (|?1+2?|?3+4? + |?1?2?|??34? ? |?1+2?|?3+4? ? |?1?2?|??34?) 2 |?1+3?|?2+4? = 1 (|?1+2?|?3+4? ? |?1?2?|??34? + |?1+2?|?3+4? ? |?1?2?|??34?) 2 |?1?3?|??24? = 1 (|?1+2?|?3+4? ? |?1?2?|??34? ? |?1+2?|?3+4? + |?1?2?|??34?) 2 Due to this change, the error rate has to be corrected as follows. On page 3 of the Article the paragraph: ?However, if she chooses incorrectly, the state she measures is random and the two qubits are not entangled, and the state sent to Bob cannot be the same as the state sent by Alice. Because there are two Bell states and every Bell state has four types (|?+?, |???, |?+?, |???), Bob gets the correct probability is 1/4 ? 1/4 = 1/16. If Bob then measures this two Bell states in the same location Alice sent, he also gets a random result, and the correct probability is also 1/4 ? 1/4 = 1/16?. Should read ?However, Eve can still obtain the correct results with the probability of 1/4 if she chooses the wrong order, while she can obtain the wrong results with the probability of 3/4. So, the correct probability that Bob gets is 1/2 ? 1/4 = 1/8. Therefore, the probability of the correct results that Bob will finally obtain is 1/2 ? 1 + 1/2 ? 1/4 = 5/8?. On page 4 of the Article the paragraph: Since Alice and Bob preserve only the part of the information that the same base they use when MEQKD protocol attacked by individual attacks, while in this part of the information, it is the probability of 1/2 to take the same base without introducing errors at this moment when Eve is eavesdropping; Simultaneously, it is the probability of 1/2 to take the different base and introducing errors with the probability of 15/32 at this moment, so the final result of the error rate is 15/32 = 46.875%. When Eve is eavesdropping, Eve gets 0 with the probability of 17/32 and gets 1 with the probability of 15/32 if Alice sends message 0; Similarly, Eve obtains 1 with the probability of 17/32 and obtains 0 with the probability of 15/32 if Alice sends message 1. Then, P (r) = ?i P (r i)P (i) = 1 , 2 P (0|0) = P (1|1) = 1372 , P (1|0) = P (0|1) = 15 , thus we can get the mutual information: 32 I (A, E) = 1 + 1372 log2 1372 + 15 15 32 log2 32 = 0.0028 Should read Alice and Bob preserve only the part of the information that they use the same base when MEQKD protocol is attacked by individual attacks. In this part of the information, there is a probability of ? of taking the same base without introducing errors when Eve is eavesdropping. Simultaneously, there is a probability of 1/2 of taking the different base, and the probability of introducing errors is 3/8 at this time. So the final result of the error rate is 3/8 = 37.5%. When Eve is eavesdropping, Eve gets 0 with the probability of 5/8 and gets 1 with the probability of (5) I (A, E) = 1 + 5 8 5 + 3 Note that the security for the QKD scheme proposed in the Article has not been proven yet. This work is licensed under a Creative Commons Attribution 4.0 International License. The images or other third party material in this article are included in the article?s Creative Commons license, unless indicated otherwise in the credit line; if the material is not included under the Creative Commons license, users will need to obtain permission from the license holder to reproduce the material. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/ 3/8 if Alice sends the message 0 . Similarly, Eve gets 1 with the probability of 5/8 and gets 0 with the probability of 3/8 if Alice sends the message 1 . Then , P (r) = ?i P (r|i)P (i) = 1 , P (0|0) = P (1|1) = 5 , P (1|0) = P (0|1) = 3 . 2 8 8 Thus, we can get the mutual information: 8 3 8 = 0 . 0456


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Jian Li, Na Li, Lei-Lei Li, Tao Wang. Corrigendum: One Step Quantum Key Distribution Based on EPR Entanglement, Scientific Reports, 2017, DOI: 10.1038/srep46201