基于广义七-粒子布朗态的量子隐形传态和量子态共享，康双勇，陈秀波，在这篇文章中,一个新颖的量子隐形传态和量子态共享方案被提出,广义的七-粒子布朗态$|B_{7}angle$被用作量子信息载体。首先,对于任意单�
国酗技论文在线 http://www.paper.edu.cn Recently, through an extensive numerical optimization procedure, Brown et al. 17 have discovered a. maximally entangled five-qubit Brown state 2(001))+010){〉+100)+11))12345 where 00)±11).y+)-一(01)±10) are Bell states. Afterwards, Muralidharan et a.l. [10 have proposed the generalized multi-qubit Brown state Bn+5 B2+=2(712100+m21010)+m2n100p++m211) (3) where na)m form a computational basis of nth order For example, if n=2, we can get the generalized seven- qubit Brown state B2(001+010104)+|10100++111+)12345 Enlightened by the works of Refs. [12, 13, 14, 15, 16, 17, we propose a novel scheine for QT and QSTs by using B7), which is a new state. B7) is different from generalized seven- particle ghz or w state under stochastic local operations and classical communication. And it optimally violates a new Bell inequality. Based on its peculiar construction. it can realize some special communication functions. In this paper, solving the problem for QT and QSTS is a new application of B7 The rest of this paper is organized as follows. In section 2, we present a perfect QT protocol and three distinct QSTS ones via. B7. Then we make an overall comparison among three QSTs ones and present an almost even distribution principle of particles. In sections 3 and 4, by mean of the principle we design several Qt and Qsts protocols of arbitrary two- and three-qubit via B7. In section 5 we consider the scheme's security. some conclusions are shown in section 6 1 QT and QSTs of a single-qubit state 1.1 QT of a single-qubit state We assume that Alice possesses an unknown single-qubit state 亚)a=(a0)+81)x where a)2+ B12=1. Alice wants to teleport y) to Bob and shares Br)with him. Initially Alice prepares the combined state of )z and qubits (1, 2, 3, 4, 5, 6 of B7), and then performs a sevell-qubit vOIl Neunann measurenent (VNM)oll these qubits. Since there are four possible 国酗技论文在线 http://www.paper.edu.cn measurement results(MRS), Alice can convey her MRS to Bob via classical channel. Depend- ing on the classical information sent by Alice, Bob can apply an appropriate paulli operator (I, ox, 0y, 02) to retrieve y)x, where I=|0)(0+11)(1,ox=|0)(11+1|1)(0, i0y=|1)0-10)(1,2=10)(0-1)(1 Thus Alice can perfectly communicate y)x with Bob via B7). This protocol is shown in detail as follows 亚x|Br) 1234567 2(a+1)(000业)+10101)+10100业)+1111+) 4三92++助)+192X-81)+(+Q113)+1-)1-18O) where Q2+)and 6+ are mutually orthogonal states of measurement basis. They are described +)=[00)(10100)-100011)+01(01001〉+11110)]123456 +[10)/(00010)-10101)+|11(01000+11111)128456 _)=100)(00010-10101))+01)(101000)+11112123456 士[10(00011)+10100)+|11)(101001)+|1110)2123456 1. 2 QSTS of a single-qubit state 45 x-1-2|345 t lirs 3 1: Different steps of protocol I K 2: Different steps of protocol II Motivated by Ref [18 we now proceed to study five QSTS protocols of v in Eq (5) imong three participants via B7), in other words, Alice wants to communicate y)a with Bob who is designated to reconstruct the original quantum information, under the help of Charlie 1.2.1 Protocol i The following steps can be illustrated the proceeding of proposed QSTs protocol I. The basic idea of this protocol is demonstrated in Fig. 1, where the solid lines connect qubits and 5 国酗技论文在线 http://www.paper.edu.cn a 1: The relation among mrs of Alice and Charlie and Bob's LuOs Alice's result Mri Charlie's result M234:6 U 00011+-0100)+|1111)a|0)+51 10100)-000112-101000)-111a|0)-11 010〉-10101)+101001)+1111)a11)+o) 1)-|01001〉-1 10100)-0001)-101000)-1111a10)+1 10100)-10001)+101000-12a10)-|1) 00010)-10101)-101001)-111)a1)+50 0010)-10101)+101001)+1111a1)-90 0101)+11+1001010101)a10)+月1 101001)+1110-100010)-10101)a0)-1) 0100941100101001+110 01000)+1111-100011)+10100)a|1)-0) 01001+1110-100010)-|10101)a|0)+|1 010021101-10010-10101)a|0)-l 01000)+1111-100011)-|10100)a|1)+|0)a 0100)+1111+10001-10100)a1)-0)-ia the dashed ones connect the measurement between qubits, they are denoted similar meanings in the following figures [SI] Distribution of qubits: Alice, Charlie and Bob possess qubits (1:(2, 3, 4, 5, 6 and (7] respectively. [Si] Alice's job: Alice first performs a Bell state measurement(BS M)on her particle pairs Le, 1. And then she tells her bSm to Bob and Charlie via classical channel [Sf Charlie s job: Charlie carries out a five-qubit vnm on his qubit pairs and tells Bob his MRS via classical channel [Sd] Bob's job: Depending on Alice and Charlie's measurement result, Bob can apply an appropriate local unitary operation(LUO on his qubits Now, let us take an example to illustrate the principle of this QsTs protocol. It is known that Alice may obtain one of four kinds of possible mrs with equal probability. Without loss of generality(WLG), firstly, we assume that Alice performs a BSM p+) and then she publicly announces mrs to Charlie via classical channel. Subsequently, Charlie should perform a five- qubit VNM on his qubits and announce his mrs to bob If Charlie s MR is 10100)-00011+01000)+11111), then Bob's qubit collapses to the state(a0)+B 1))7, It is exactly V)z and Bob needs to do nothing o if Charlie's MR is 10100)-00011-01000)-11111, then Bob's qubit is projected onto 0-B1)7, then Bob can reconstruct vx by performing operation on his qubit 6 国酗技论文在线 http://www.paper.edu.cn Moreover, we can find that this stage requires the transmission of six (two in the SI and four in the Si)bits of classical information(chits) between the parties in total. The relation among Alice's and Charlie's MRS and LUOs with which Bob reconstructs yx is shown in Tab. 1, where M1 and M23456 denote Mrs of Alice and Charlie. The normalization factors are omitted for convenience. In order to avoid unnecessary repetition we leave out the illustrates of some symbols and normalization factors in the following figures and tables 图3: Diffe of protocol III 1.2.2 Protocols ii and iii The basic ideas of protocols II and IIi are demonstrated in Figs. 2 and 3, respectively There also exists four steps for illustrating the proceeding of protocols II and li. These steps are the same as the proposed steps in protocol T except the fist step(sad: distribution of qubits For example, Alice, Charlie and Bob possess qubits (1, 2. 3, 4, 5, 6 and(7 in protocol II, while qubits ( 1, 2, 3,4,5, 6 and (7 belong to Alice, Bob and Charlie in protocol III. We can make the corresponding operation in other steps. Additiona lly, the classica l consume are six and five chits for protocol II and IIl, respectively. The relation among mrS of Alice and Charlie and LUOs with which Bob reconstructs y)2 is shown in Tabs. 2 and 3, respectively 2.3 Comparison and discussion For an arbitrary single-qubit, state, there are another two QSTS protocols via B7,, but, we dont listed them one by one and only simply add to Figs. 4 and 5, which depicts their basic ideas. The proceedings of Figs. 4 and 5 are analogous as Figs. 1 and 2, respectively. Thus Bri can be used for QSTS of an arbitrary single-qubit state perfectly. What's more, it could also be used for QSts of n-qubit G Hz type st n by ghain=a0)on +B 1on. Through further analysis, we can see that the classical consume are different in aforemen tioned three protocols. One requires 6 chits in total both in protocol i and I. while only 5 cits in protocol II. In the view of resource saving protocol Iii may be optimal among three protocols though it utilizes more quantum resources. However, in the realistic scenario, 国酗技论文在线 http://www.paper.edu.cn E 2: The relation among mrs of Alice and Charlie and Bob's LuOs Aice’ s result m12 Charlies result 113456 y〉r 000+01)+10)+11)10100)-1001)+1000)+111a0)+B1) 000+1001)+|10)+111)10100)-10011)-1000}-111)a|0)-B1)a 000+-10012+110)+1110010)-10101)+1001)+11)0a1)+B|0)a 100+101:+110+-111001)-10101)-1012-111041)-0)-27n 0o+1o)-10)-11)10100)-10011)-1000)-11)al0)+1 009+1001)-10)-11)10100)-1001)+1000+111alo)-B|1 000)+1001)-110 |10010)-10101)-1001)-10)a1)+0)am 000+1001)-|10)-11110010)-10101)+1001)+111)a1)-B|0)-10 010)+1011+100+101)1001+110)+0010)-10101)0)+1)7 010)+101)+100)+101)1001)+10-10010+10101)c0)-|1) 010)+10112+100)+101)1000+111-0011+10100)a1)+|0) 010}+1011)+100)+101)1000+111)+1001)-10100)a1)-|0) 010}+1011)-100)-101)1002+110)-0010)+10101)a|0)+1) o10)+10112-100-101)1001)+110+10010-10101)a|0)-1 010)+101)-100-101)1000+1114001-10100a1)+510)a 010)+1011-100)-101)1000+11-0011+10100)a1)-0)-i0g very difficult for protocol Iii to perform Alice's six-qubit VNM. Thus it's not feasible experi mentally. At the same time, for protocol I, it may encounter the same trouble as protocol III Furthermore, in protocol II mrs of Alice and Charlie can be realized through a series of simple operations such as BSM and single-qubit measurement Consequently, it shows that protocol II is the Inost optimal owing to reasonable distri bution of particles. It refers to the number of system qubits excluding receiver's should be even distribution or the quantity difference of qubits is as less as possible between sender and collaborators. We name it as almost even distribution principle of particles 2 QT and QSTS of an arbitrary two-qubit state 2.1 QT of an arbitrary two-qubit state For teleporting an arbitrary two-qubit state xg=(ao00)+a101)+a210)+a311)y 0 via Br), where iso la: 12=1. Alice can obtain a nine-qubit state and then makes a seven-qubit VNM using orthogonal states shown in the first column of Tab. 4, Alice sends her mrs to Bob via classical channel. Subsequently, Bob applies a suitable luo to recover y)ay. Hence, QT of an arbitrary two-qubit state is deterministically implemented. The relation between Alices MRS and the corresponding LUOs is shown in Tab. 4, where Meul2345 and U6 &U represent Alice's MRS and Bob's LUOs on their particles, respectively 国酗技论文在线 http://www.paper.edu.cn k3: The relation among mrs of Alice and Charlie and Bob's LUOs Alice's result M=12345 Charlies result MG 2 r t C1+|101010) a|1)+60) C00001)+|101010) al10)+B|1)1 000001)-1101010) a|1)-B|0)|-iy 00001-101010) 1)-a|0) 000001)+111111 1)+β|0 000U1)+|1111) 000001)-|11111 Lo y 000001〉-111111) c10)+B11)I 010100)+10101a) 1)+0) 010100)+|101010) a|0)-|1) 010100)-1101010 1〉-B|0 010100)-1101010) 0)+B1)1 010100)+|111111) a|1)+B0) 010100)+111111 l)+1)|1 0100)-111) 0 |1)-B0) 010100)-11111) 1 0〉-B1) 1U001)+1000 c|0)+B1) 001010)+|10001) 1 a|1)+30) 001010)-110001) a|0)-B1 01010)-|100001) 001010)+|110100 c|0)+B|1 0O1010)+|110100) B 0)-c1-ia UU1010)-1110100) a|0)-B|1) 001010〉-110100 a|1)+0) 0111110001 al1)+B1) 011111+1100001) c1)-B|0 011111-100001) 0)-Bl1)|a 011111}-|100001) a|1)+B|0 0111112+1110100) a|0)+61 0111112+1110100) a|1)+B|0) 1111)-110100 011111}-|110100) 1)-B|0)〉|-ic 9 国酗技论文在线 http://www.paper.edu.cn Now we proceed to study three QSTS protocols of y xy among three participants via B7) charlo R 4: This is the similar as Fig. 1 k 5: This is the similar as Fig. 2 2.2 QSTS of an arbitrary two-qubit state In order to concise our texts and avoid repetition, we dont list three Qsts protocols one by one, but Figs. 6 and 8 call adequately reflect the basic ideas of theIn. Conbining with our almost even distribution principle of particles we only choose Fig. 6 to illustrate the QSTs protocol of y)zy. Furthermore we could find it is the most optimal. The steps can be described as follows [S Distribution of qubits: Alice, Charlie and Bob possess qubits ( 1, 2, 3, 1, 7 and 15, 61 respectively So Alice's job: Alice first performs a four-qubit VNM on her particles. And then she tells her bsm to bob and charlie via classical channel [Si Charlie's job: Charlie carries out aa three-qubit VNM on his qubit pairs and tells Bob his mrs via classical channel [ Sd Bob's job: Depending on Alice and Charlie's measurement result, Bob can apply an appropriate local unitary operation(LUO) on his qubits The relation among mrs of Alice and Charlie and LUOs with which Bob reconstructs xy is shown in Tab. 5 10 国酗技论文在线 http://www.paper.edu.cn ( 6: Different steps for QSTS of an arbi- K 7: Different steps for QSrS of an arbi- trary two-qubit state via B7) trary three-qubit state via B7) 叶出」 3 8: Different steps for another two QSTS protocols of an arbitrary two-qubit state via B7 3 Qt and Qsts of an arbitrary three-qubit state 3.1 QT of an arbitrary three-qubit state For teleporting an arbitrary three-qubit state 业Dy2-(a000+a1001+a2010+011+a410+a5101+a110+a7111)2(11) via B7), where 2i-olaiI2=l. Alice can obtain a ten-qubit state and then makes a seven-qubit VNM, then Alice sends her Mrs to Bob via classical channel. Subsequently, Bob applies an suitable luo to recover y)ry2 Let us take an example to demonstrate the principle of QT WlG, we assume Alice obtains the state 0011111+1011011-0111101-1111001 and then she sends her mrS to Bob via four chits. Bob's qubits collapse to the state uol011)+a11ll-u21000)-a3 100)-W4001)