论文研究-The general theory of three-party quantum secret sharing protocols under phase-damping channels.pdf

所需积分/C币:5 2019-08-16 254KB .PDF
评分

相位阻尼信道下三方QSS协议的一般性理论分析,宋婷婷,秦素娟,本文讨论了噪声信道下三方QSS协议的一般性理论。对于初始三粒子三体纯纠缠态,某一个粒子在噪声信道传输后,系统三粒子态变为混合态�
国利技论文在线 http://www.paper.edu.cn each other can easily recover the secret, even though the states are changed into mixed states after transmitted through phase-damping channels. For the Qss protocol based on GHz-class states under the phase-damping channels, the method for recovering the secret is hard because of the difficulty discrimination of the two ultimate mixed states. However, we find out the optimum unambiguous discrimination with the maximum confidence and the minimum failure probability. Furthermore, we analyze the security of one recent Qss protocol with noisy chan- nels[ 11, and conclude that the protocol does not satisfy the sa fety standards of Qss protocols By the correlation-extractability of gHZ states, any illegal participant could get more than one half of Alice's secret without the help of others, and the attack strategy cannot be detected The structure of this paper is as follows. Sec. 1 is about the Qss protocol based on the w-class states under phase-damping channels. The Qss protocol based on the GHz-class states is shown in Sec. 2, and the participant attack strategy on the recent QSs protocol is also figured out in this section. The conclusion is summarized in sec. 3 1 The QSs protocol based on W-class states under the phase-damping channels In this protocol, the quantum channel is phase-damping channel, which means that the channel describes the loss of quantum information without loss of energy. And the operations of a phase-damping channel can be given by three Kraus operators Eo-vI-pI, E1-ypl0y(0 and E2=Vpl1(, where p, a real number in the set [0, 1, is the channel's parameter. The three parties are denoted by alice, Bob and charlie. Bob and Charlie want to work together to share the classical secret of Alice. The secret is randomly extended into one of the message 101, 10, and the secret "1"is also randoMly changed into one of the message 100, 11]. So the length of message is twice the length of her secret. Before beginning the protocol, alice prepares a string of W states, and each state is V)ABC=1/v3(1001)+ 010)+|100))ABC Particles A, particles B and particles C form three sequences, denoted by sequence A, sequence B and sequence C respectively. The process of three-party QSs protocol based on w-class states under the phase-damping channel is showed as follows 1. Alice encodes each of the two-bit message on the two-bit unitary operation pair (UBOUc which is operated on the pair of particle b and particle C. Both of UB and Uc are from the Pauli operators I ,0x, 0x, io, The detailed encoding strategy [12 and the three-qubit system after operated y'ABc are showed in Table 1 2. Alice inserts some decoy states into the sequence B and sequence C, which are random from the set (0),1),1+)=1/v2(10)+ 1)),1-=1/v2(0)- 1))). The final sequences are denoted by Band C, respectively. Then Alice holds the sequence A, and sends the 国利技论文在线 http://www.paper.edu.cn W 1: The encoding strategy and the states of yaBc Message Unitary operation State of YABc) 0 16I or a280。1/√3(100±1001)±010) 01 I⑧aor2⑧ioy1/√3(±101)+1000+011) o2②a2 or 2078I1/√3(±110)+1000)-011) 11 Froyo 2o⑧a1/3(1112千001)±010)) sequence Band Cto Bob and Charlie through the phase-damping channels, respectivel 3. Bob and Charlie receive the sequences, and then Alice announces the positions and the corresponding bases of decoy states in sequences B and C. Due to the phase-damping noise in channel, the sequences B and C are affected even there is no eavesdropper When the decoy state in sequences B' (C")is modulated into the states 0), 11, Bob ( Charlie) also measures the state with the rectilinear basis ( 0,1)), because the states in basis (0), 1) will not be changed under the phase-daInpinlg channel. Whell the decoy state is in the diagonal basis (+),1-, Bob(Charlie) operates the maximum confidence measurement,and the measurement operators are Il1 = ++1: I2=--. If the measurement result is+>, Bob(Charlie)denotes the classical bit 0", otherwise Bob (Charlie) denotes the classical bit "1". Under the phase-damping channels, the states +) and the states - will become(1-2)+(++2- and(1-2)-)(1+21+(+, respectively. So the phase-damping channel would introduce the errors with the proba. bility p/2 for diagonal basis. Bob and Charlie public their measurement results of the decoy states, respectively. Alice calculates the error rates. If the error rate is below the threshold value, they proceed; otherwise, they re-execute the protocol 4. According to the each message. the three-particle state is changed into the following mixed 国利技论文在线 http://www.paper.edu.cn *2: The states and the probabilities after Alice measures her qubits Message Alice's Corresponding State shared by Bob and Charlie result probabilit /3 BC 1 1/3 0000 0 2/3 1 1/3 01)/(01 1/3 10)/10 0 /3 3 11/(11 px:y=2(01)(011+10(10)=2(01)o+10)(O1D (00)(00|+11)(11)= (1 Po1(10) (00)(111+|11)001) entangled state because of the phase-damping noise ABc 3(1000010+-3 1-p)2(0001-0100101 1P(士10001+100100100+1010(100) ABC (1-p) (101)(1011+100090001+011)(011)+ 000)(011+011)(000 3(+10)(00000+10)1+01(101) A弓 p10 11010+1000001002-(0000000 32(10101091010090 ABC (111009001+1010010(1-p)2(010)01+00)010 -。(士010)(111±1129010千00)(111千11119001) (1) 5. Alice Ineasures the qubits in her hand with the rectilinear basis (0), 1) aild announces the measurement results to Bob and Charlie. If the state is 0), Alice announces"0"to Bob and Charlie; otherwise, she publics"1". Table 2 gives the Alice's measurement results and the corresponding states shared by Bob and Charlie 6. When Bob(Charlie) wants to recover the Alice's secret, he needs Charlie's(Bob's) help 国利技论文在线 http://www.paper.edu.cn RE 3: The states and the probabilities after Charlie(Bob) neasures his qubits Alice's Charlie's(Bob's) Corresponding Bob's(Charlie's) Message result result probability state 1/2 00 2 0)(0 3 01 0)(0 1/2 10 2 1)(1 1/2 11 1)(1 1)(1 p1(4=1210)(01+|a21)(11=(1-p)2aB(|0)(11+1)(o) p23)=|a|2|0)(0+1921)(11=(1-p)2aB(|0)(1+|1)(O) Suppose Charlie(Bob)measures his qubits with the basis do=a0)+B 1, p) B 1f, where a and B satisfy the conditions a 2+B 2=1 and aB*=a*B. Charlie (Bob) denotes the measurement results o)andlφ?)by“o”and“1” respectively.Then the measurement results are sent to Bob(Charlie) through the public classical channel Now the states and the corresponding probabilities are changed into the cases shown in Table 3 7. Only when a|=0 or B=0 holds, Bob( Charlie), together with Charlie(Bob),can correctly recover the Alice's secret. Actually, when a=0 holds. Charlie firstly Hips his bit values. In both cases, the measurement operators used by Charlie(Bob) in last step are 0(0, 11. Bob( Charlie) measures his bit sequences with rectilinear basis and obtains his bit values. Then Bob can recover each bit of Alice's secret by performing the bit-XORs operations on his result. Alice's result and Charlie's result. The particular 国利技论文在线 http://www.paper.edu.cn 4: The recovered process of final key Message Alice’ s result Bob’ s result⊕ Charlie’ s result Key 0 0 00001 results are in table 4 Now we analyze the security of the QSs protocol. For external eavesdroppers, they cannot obtain any information without introducing the errors, because the bbs4 decoy particles can keep each transmission secure. For the untrusted participant, he cannot get any information without the other's help. In step 4, the state hold by Bob is the maximum mixed state (0(0+11/2(Here, we suppose the probability of0"in Alice's secret is the same with that of“0”). Moreover,it' s the maximum mixed state when Alice encodes secret“0”.i.e., when the message Is“01”or“10”. When the message are“00”and“11, the secret is“1”,Bob’ s state is also the InaxiInuIn Inixed state. Thus, Bob could not get ally informlation about Alice's secret only from the particle hold by himself. By the same analysis, it's easily known that Charlie could not get any information about Alice's secret only from the particle hold by himself, either Furthermore, after step 5, after Alice publics her measurement result, Bob's state and Charlies state are the maximum mixed state, respectively, for both the secret“0”and“1”. They could not get any information about the secret according to the state hold by themselves without the others help. Thus our scheme is immune to the participant attacks. The scheme based on W-state under phase-damping channel is secure. This special encoding strategy could cover an orthonorma l basis of three-qu bit states which has 8 orthogona. I w states. Actually if Alice uses only one unitary operation pair for each message, this protocol is also secure 国利技论文在线 http://www.paper.edu.cn 2 The QSs protocol based on GHz states under the phase-damping channel For GHz-class states, we also could give a secure QSs protocol under the phase-dailping channel. The effect of phase-damping channel is represented by the three Kraus operators Eo=v1-pI, E1= VplO(0 and E2=pll>(1, where p is a real number in the set [0,1] this protocol, the initial states are GHZ states, which are D)aBc-1/v 2(1000)+(111)).After some particles are transmitted through the phase-damping channel, the states are changed into mixed entangled states 2.1 The three-party QSS protocol based on GHz states under the phase damping channel Consider three parties, Alice, Bob and Charlie. Alice want to share her secret between Bob and Charlie. The encoding strategy and the recovering process are shown in the following steps 1. Alice prepares the GHZ states )ABC=(000)+|111)/v2, keeps the first-qubit sequence A, and sends the second-qubit sequence B, third-qubit sequence C through the phase- danping channels to Bob and Charlie, respectively. Before the sequences are sent to bob and Charlie, in order to make sure the sequences are transmitted securely, they can be inserted some decoy particles, which are random chooses from the sets 0,1,+> 1/2(0)+1).-)=1/V2(0)-1)} 2. After Bob and Charlie receive the sequences, Alice publics the positions and the bases of the decoy states. The phase-damping channels only introduce errors for the states in diagonal basis, and the error rate is p/2. Considering this, Bob and Charlie measure the decoy states with the corresponding bases, and announce the measurement results Alice estimates the error rates. and the error rates should be below the threshold values otherwise they will re-execute the protocol. Each aBc is transformed into pABC=(000900+111(111 (00011+111900) 3. Bob encodes two bits of information by performing the Pauli operators ((00), 02(01) ig, (10), 02 (11) on each particle of the sequence B. At the same time, Charlie also encodes his two bits of information by carrying the Pauli operators (I(00), ox(01), io,(10) o2(11)) on each qubit in sequence C. Then PABC is encoded into Pii=IQUiQUPABcIO Vi& Uj, where i,3E 1,x,,2 and UiG represents one of the four Pauli operations I1, 0x, iov, 021. After performing the unitary operations, the sequences B and C are sent back to Alice by inserting some decoy states in both sequences 国利技论文在线 http://www.paper.edu.cn 4. Alice makes sure that there is no eavesdropper in both of the channels. With the ghz basis, she measures the corresponding three qubits which is in Eq.(2), because these measurement operators maximizes the successful probability that Bob and Charlie could recover the secret. The particular reason is shown in Appendix A. Then Alice records the measurement results according to table 5. and the results are her secret ZZ (000900+111)111) (000(11+1119000) p1x-p2zy-(00901+1010)+ (1-p)2(0010+1001 py=pzx=(001)0011+10)(110) (001)(110+1110)(001) pIz=zr=,(000011 (1-p) 2(000(1111900 (1-p)4 pxr=pyz=2(010)(010+101)101)+ (010)(101+1101)(010) Pxx- Py (011)(011+1100)(100) 2012(100|+1009011 011)(011+|100)(100) (1-p) (011)(100|+100)(011 0xz==(010)010+101)101)-2,-(0(101+101)010)(2) N 5: The secret corresponding to the measurenent result (000+111)/V2(001+110)2(001)-11)/V2(000-111)/ Meaurement result (011+|100)/y2(010)+101)/2(010)-101)/V2(011)-100/2 Alice's secret 10 5. Bob and Charlie want to cooperate with each other to recover the secret of Alice at some time. If Bob recovers the secrets, Charlie tells his encoding information to Bob Bob operates the bit-XOR operations on his bits and Charlie's bits, and the results are recovered information. According to Appendix A, we can know the probability that the results recovered by Bob and Charlie are coincident with Alice's secrets is [1+(1-p)4/2 Thus Bob and Charlie could recover Alice's secret except the probability 1 p)4/2 The security of this Qss protocol can be analyzed easily, because the security of the quantulnl coMmunications through channel is checked by decoy states. External eavesdroppers could not get any in formation while the particles are transmitted through the phase-damping 国利技论文在线 http://www.paper.edu.cn 1: The maximum confidence C(solid line) and the error rate P(dashed line) with the parameter p of phase-damping channel channel. Moreover, after step 2, there is no information leaked to untrusted parties, which is guaranteed by the encoding strategy. The particular analysis is as follows. In our protocol ach state in sequence B and sequence C is the maximum mixed state, and this cannot leak any information to Bob and Charlie. After Bob and Charlie encode the information, each particle in sequence B and sequence C are randomly in 0) and 1). Thus Bob(Charlie)also could not get any information from the quits. While the qubits are sent back to Alice through the phase-damping channel, the security of qubits are a so guaranteed by decoy states. So our protocol is secure. Moreover, the successful probability is obviously related with the parameter of phase-damping channel p. Figure 1 depicts the maximum confidence and the error rate with the channel parameter p as the variable. As we can see, the confidence declines as the channel noise p increasing, and to the contrary, the error rate gets larger with the channel noise increasing. However, both the confidence and the error rate approach to 0.5 with the p close to 1. This is because Bob and Charlie could randomly guess Alice's secret when the channels have large noise, which is the worst case for the Qss protocol. All of them are consistent with the practice. However, the secret shared by parties is a little smaller than that of the protocol proposed recently by Adhikari et al 1l(i.e, ACA), which is also larger than 1. The comparison is shown in Fig. 2, and the ACAs protocol also considers the security of QSs protocol under the phase-damping channel. Furthermore, we find that this scheme doesnt satisfy the safety criteria of QSs protocols, and the ACas protocol could leak one bit secret to che illegal participants. In the following we will give the detailed description about the acas protocol

...展开详情
立即下载 最低0.43元/次 学生认证VIP会员7折
举报 举报 收藏 收藏
分享
3.68MB
protoc-gen-go和protoc.zip

protoc.exe protoc-gen-go.exe 生成protobuf的序列化和反序列化文件 命令:protoc.exe --plugin=protoc-gen-go=./protoc-gen-go.exe --go_out=./ your.proto

2019-05-27
10.75MB
开源项目-lyft-protoc-gen-validate.zip

开源项目-lyft-protoc-gen-validate.zip,用于生成polyglot消息验证器的Protoc插件

2019-09-03
4.02MB
protoc-gen-go

goprotobuf 提供的 Protobuf 插件 protoc-gen-go(被放置于 $GOPATH/bin 下,$GOPATH/bin 应该被加入 PATH 环境变量,以便 protoc 能够找到 protoc-gen-go) 此插件被 protoc 使用,用于编译 .proto 文件为 Golang 源文件,通过此源文件可以使用定义在 .proto 文件中的消息。

2018-06-29
4.01MB
protoc-gen-go.exe

protoc-gen-go.exe go golang protoc 工具,下载即用,

2018-03-28
1.5MB
protoc+protoc-gen-grpc-java

在java下使用gPRC需要编译对应的proto文件,本资源提供的两个文件用于编译proto文件生成对应的序列化用代码和通信用代码。protoc-gen-grpc-java版本为0.13.2

2018-08-15
2.17MB
protoc-gen-go.exe protoc.exe

protoc-gen-go.exe protoc.exe 用于对protobuf文件进行go语言编译,

2018-11-23
6.52MB
protoc-gen-lua-master 整合所需资源

lua-5.3.4_Win64_bin.zip protobuf-python-3.0.0.zip protoc-3.0.0-win32.zip protoc-gen-lua-master.zip

2018-02-26
12.12MB
ipmi-second-gen-interface-spec-v2-rev1-1.pdf

IPMI2.0 spec,目前BMC的规范手册,对于学习、测试BMC很有必要。

2016-08-15
27KB
protoc-gen-lua

protoc-gen-lua是protobuf的工具。官方正品。下载下来之后,文件名改一下,把“-master”去掉,是为了看我教程方便理解。

2015-11-18
104KB
protoc-gen-doc, Google协议缓冲区的文档生成器插件.zip

protoc-gen-doc, Google协议缓冲区的文档生成器插件 protoc-gen-doc 这是一个用于Google协议缓冲区编译器( protoc )的文档生成器插件。 插件可以从 .proto 文件中的注释生成 HTML 。JSON 。DocBook和 Markdown 文档。它支持pr

2019-09-18
263KB
excel-gen.js

导出excel的jquery代码,适用asp、html、jsp等,浏览器ie8+以上

2018-05-15
50.76MB
protoc-gen-lua-master proto生成lua

有问题可以给我留言,我自己项目也再用这个,python特别不好配置

2017-12-15
22.72MB
ipmi-second-gen-interface-spec-v2-rev1-1,一共600多页的文件,绝对是一手最好的资料

ipmi-second-gen-interface-spec-v2-rev1-1,因特尔提供的2020年最新资料,一共600多页的文件,绝对是一手最好的资料,看完这个,你就不用看国内摘抄的了,也不要看别的资料了

2020-04-06
2.41MB
second-gen-interface-spec-v2-rev1-4

IPMI2.0规范,适用于BIOS开发、BMC开发工作者,以及其他用到IPMI规范的工作者

2018-04-30
3.57MB
midas-gen 内部资料

midas-gen 内部资料

2014-12-10
535KB
论文研究-EPCglobal Class 1 Gen 2标准的RFID高效双向认证协议.pdf

提出用插值函数来计算Log-MAP算法中的校正函数,并在AWGN信道上采用分段差值方法实现了Turbo译码。该算法解决了校正函数计算复杂度较大的问题,消除了译码计算中的指数和对数运算。仿真结果表明:用2段二次样条函数实现的Turbo译码器,其译码性能与Log-MAP算法等价,而计算的复杂度明显降低,运算时间大幅度减少。

2019-09-10
19.5MB
NGUI Next-Gen UI v3.7.9

NGUI不解释(NGUI Next-Gen UI v3.7.9),支持Unity5.X(仅供学习使用)

2018-12-21
12.63MB
NGUI Next-Gen UI v2018.3.0e.unitypackage

NGUI Next-Gen UI v2018.3.0e.unitypackage 不多解释,懂的自然懂,亲测可用。 请低调分享。

2019-03-06
36.76MB
Twitter手机端安装包--Android

Android手机Twitter客户端,很多时候下载特别慢,希望对你有帮助。

2017-09-29
1.5MB
60分钟学会OrCAD-Capture-CIS

60分钟学会OrCAD-Capture-CIS 很不错的资料,推荐给大家

2017-09-29
img

关注 私信 TA的资源

上传资源赚积分,得勋章
相关内容推荐