Page 1

Schemes of Teleportation with Six-qubit Cluster State Qian Lan*1, Xinwei Zha2, Jing Wei3 School of Science, Xi’an University of Posts and Telecommunications, Xi’an 710061, China lanqian198711@163.com; 2zhxw@xupt.edu.cn; 3weijing129@126.com

*1

Abstract In this paper, we have proposed a quantum state teleportation

of

an

arbitrary

two-qubit

state

using

non-maximally entangled six-qubit Cluster state as channel. The first scheme is described as: the sender Alice introduced an auxiliary particle and operators orthogonal complete bases measurement, and the controller Bob performed non-Bell measurements on his particles in the quantum

Cluster state as channel. In the first scheme, the sender Alice first introduces an auxiliary particle and performs an orthogonal complete bases measurement; the second one is the receiver introduces an auxiliary particle and performs an appropriate unitary transformation, the unknown particle state can be teleported. And both of the two schemes’ probability 2

of success is proved to be 4 a 。

teleportation process. At last, the receiver Charlie can get the arbitrary two-qubit state only by operating appropriate unitary transformation. The second scheme is: the sender

The First Telepertation Arbitrary Two-Qubit State

Scheme

of

the

Alice operated joint Bell state measurement, and the controller Bob performed non-Bell measurements on his

Suppose that the sender Alice plans to teleportate the

particles, the receiver introduced an auxiliary particle and

following arbitrary two-qubit entangled state to

operated appropriate unitary transformation, the quantum

Charlie

|   a1a2  ( x0 00  x1 01

teleportation will be successful.

 x2 10  x3 11 ) a1a2

Keywords Quantum Teleportation; Non-Bell Measurements; Joint Bell State Measurement; Unitary Transformation

Where

x0 , x1 , x2 , x3

(1)

are arbitrary complex numbers,

and it is assumed that the wave function satisfies the Introduction Quantum communication is a popular topic in the communication field. The quantum communication includes many sides, such as quantum teleportation [1-14], quantum key distribution [15], quantum secure direct communication [16], dense coding [17-19], quantum secret sharing [20], remote state preparation [21-22] and Quantum state sharing [23-24] which is also named QSTS. QSTS, which plays an important role in quantum communication, is the scheme of sharing an unknown state among some agents. Recently Cluster state has been discussed a lot not only in quantum information but also in quantum communication. In this paper we compared two schemes of teleporting an arbitrary two-qubit state which used a non-maximally entangled six-qubit

JOURNAL OF OPTICS APPLICATIONS – Oct. 2012, Vol. 1, Iss. 2

normalization condition

2

3 i 0

xi

 1 .The six-qubit

cluster state can be written as

C6

A1 A2 B1 B2 C1C2

 (a 000000  b 010101  c 101010  d 111111 ) A1 A2 B1B2C1C2

And

(2)

a  b  c  d, a  b  c  d  1 . N o w w e 2

2

2

2

consider that Alice, Bob, and Charlie hold particles ( A1 , A2 ), ( B1 , B2 ), ( C1 , C2 ) respectively. In order to teleportate the state

a1a2

in this scheme, Alice first

introduces an auxiliary particle 0 A .The system state of particles becomes

14


s

|   a1a2  C

A1 A2 B1 B2 C1C2

 0

2

1 a a a g8  ( 00001  1    00011  01000  10101 2 b d b

A

 [ x0 (a 000000000  b 000010101

2

a a a  1    10111  11100  1    11110 ) a1a2 A1 AA2 c d  c

 c 001001010  d 001011111 )  x1 (a 010000000  b 010010101

(3)

 c 011001010  d 011011111 )  x2 (a 100000000  b 100010101

 c 111001010  d 111011111 )]a1a2 A1 AA2 B1B2C1C2

2

(4.9)

2

1 a a a a g10  ( 00100  1    00110  01101  1    01111 2 c d c d 

step1: In order to realize the teleportation, Alice can make a measurement on the five particles ‘ a1a2 A1 AA2 ’ a nd co n vey her r es ul ts to Bo b vi a cl a ssi c a l communication. If Alice operates a measurement using i a1a2 A1 AA2

2

a a  10000  11001  1    11011 ) a1a2 A1 AA2 b b

 x3 (a 110000000  b 110010101

g

2

1 a a a a g9  ( 00100  1    00110  01101  1    01111 2 c d c d  2

 c 101001010  d 101011111 )

the states

(4.8)

2

, i  1, 2,3,...30,31,32 , which are

(4.10)

2

a a  10000  11001  1    11011 ) a1a2 A1 AA2 b b 2

2

1 a a a a g11  ( 00100  1    00110  01101  1    01111 2 c d c d 

(4.11)

2

a a  10000  11001  1    11011 )a1a2 A1 AA2 b b

given as follow: 2

2

1 a a a g1  ( 00000  01001  1    01011  10100 2 b c b 2

(4.1)

2

a a a  1    10110  11101 1    11111 ) a1a2 A1 AA2 d c d 

2

1 a a a a g12  ( 00100  1    00110  01101  1    01111 2 c d c d a a  10000  11001  1    11011 ) a1a2 A1 AA2 b b 2

2

1 a a a g 2  ( 00000  01001  1    01011  10100 2 b b c   2

(4.2)

(4.3)

2

2

1 a a a g 4  ( 00000  01001  1    01011  10100 2 b c b

(4.4)

2

2

(4.5)

2

(4.6)

2

2

a a a  1    10111  11100  1    11110 ) a1a2 A1 AA2 c d  c

15

2 2 1 a a a g17  ( 00010  1    01001  01011  1    10100 2 b b c

(4.17)

2 a a a  10110  1    11101  11111 ) a a A AA 12 1 2 c d d  

2

a a a  1    10111  11100  1    11110 ) a1a2 A1 AA2 c d  c

2

(4.16)

a a  10001  1    10011  11000 ) a1a2 A1 AA2 b b

a a a  1    10111  11100  1    11110 ) a1a2 A1 AA2 c d  c

1 a a a g7  ( 00001  1    00011  01000  10101 2 b d b

2

1 a a a a g16  ( 00101  1    00111  01100  1    01110 2 d c d  c 2

2

1 a a a g6  ( 00001  1    00011  01000  10101 2 b d b

(4.15)

a a  10001  1    10011  11000 )a1a2 A1 AA2 b b

a a a  1    10110  11101  1    11111 ) a1a2 A1 AA2 d c d 

1 a a a g5  ( 00001  1    00011  01000  10101 2 b d b

2

1 a a a a g15  ( 00101  1    00111  01100  1    01110 2 d c d  c 2

2

2

(4.14)

a a  10001  1    10011  11000 ) a1a2 A1 AA2 b b

2

2

2

1 a a a a g14  ( 00101  1    00111  01100  1    01110 2 d c d  c 2

a a a  1    10110  11101  1    11111 ) a1a2 A1 AA2 d c d 

2

(4.13)

2

2

2

2

2

1 a a a a g13  ( 00101  1    00111  01100  1    01110 2 d c d  c a a  10001  1    10011  11000 ) a1a2 A1 AA2 b b

2

a a a  1    10110  11101  1    11111 ) a1a2 A1 AA2 d c d  1 a a a g3  ( 00000  01001  1    01011  10100 2 b c b

(4.12)

2

(4.7)

2 2 1 a a a g18  ( 00010  1    01001  01011  1    10100 2 b b c

(4.18)

2 a a a  10110  1    11101  11111 )a1a2 A1 AA2 c d d 

JOURNAL OF OPTICS APPLICATIONS – Oct. 2012, Vol. 1, Iss. 2


2 2 1 a a a g19  ( 00010  1    01001  01011  1    10100 2 b b c

2

(4.19)

a a  1    10001  10011  11010 ) a1a2 A1 AA2 b b 2

2

1 a a a a g30  ( 1    00101  00111  1    01100  01110 2 d d c c    

(4.20)

2 a a a  10110  1    11101  11111 ) a1a2 A1 AA2 c d d 

2 2 1 a a a g21  ( 1    00001  00011  01010  1    10101 2 b b d 

(4.30)

2

a a  1    10001  10011  11010 ) a1a2 A1 AA2 b b   2

2

1 a a a a g31  ( 1    00101  00111  1    01100  01110 2 d c d  c

(4.21)

2 a a a  10111  1    11100  11110 ) a1a2 A1 AA2 d c c

(4.31)

2

a a  1    10001  10011  11010 ) a1a2 A1 AA2 b b 2

2 2 1 a a a g22  ( 1    00001  00011  01010  1    10101 2 b b d 

(4.29)

2

2 a a a  10110  1    11101  11111 ) a1a2 A1 AA2 c d d  

2 2 1 a a a g20  ( 00010  1    01001  01011  1    10100 2 b b c    

2

1 a a a a g29  ( 1    00101  00111  1    01100  01110 2 d c d  c

2

1 a a a a g32  ( 1    00101  00111  1    01100  01110 2 d c d  c

(4.22)

(4.32)

2

a a  1    10001  10011  11010 ) a1a2 A1 AA2 b b  

2 a a a  10111  1    11100  11110 ) a1a2 A1 AA2 d c c

If Alice’s measurement outcomes are: 2 2 1 a a a g23  ( 1    00001  00011  01010  1    10101 2 b b d 

{ gi

The unknown particles entangled state can be teleportated, if Alice’s measurement outcomes are:

gi g24

(4.24)

2 a a a  10111  1    11100  11110 ) a1a2 A1 AA2 d c c 2

2

1 a a a a g25  ( 1    00100  00110  1    01101  01111 2 c d c d 

, i  1, 2,3,...14,15,16 }.

(4.23)

2 a a a  10111  1    11100  11110 ) a1a2 A1 AA2 d c c  

2 2 1 a a a  ( 1    00001  00011  01010  1    10101 2 b b d 

a1a2 A1 AA2

a1a2 A1 AA2

, i  17,18,19,...30,31,32 .

The unknown particles entangled state can not be teleportated.

Therefore,

outcome is{ gi

a1a2 A1 AA2

if

Alice’s

measurement

, i  1,2,3,...14,15,16 }.

Alice can inform Bob of the measurement outcomes via (4.25)

a classical channel.

2

a a  10010  1    11001  11011 ) a1a2 A1 AA2 b b 2

Step2:

a1a2 A1 AA2

,

the state of particle B1 B2C1C2 will collapse into the

2

1 a a a a g26  ( 1    00100  00110  1    01101  01111 2 c c d d    

If Alice’s measurement outcome is g1

(4.26)

following states:

2

a a  10010  1    11001  11011 ) a1a2 A1 AA2 b b 2

2

1 a a a a g27  ( 1    00100  00110  1    01101  01111 2 c d c d 

(4.27)

2

a a  10010  1    11001  11011 ) a1a2 A1 AA2 b b   2

2

1 a a a a g28  ( 1    00100  00110  1    01101  01111 2 c d c d 

(4.28)

2

a a  10010  1    11001  11011 ) a1a2 A1 AA2 b b

JOURNAL OF OPTICS APPLICATIONS – Oct. 2012, Vol. 1, Iss. 2

B1 B2 C1C2

a ( x0 0000  x1 0101 2  x2 1010  x3 1111 ) B1B2C1C2

(5)

Then Bob takes a joint X basis measurement Xˆ B1  Xˆ B2 on his two particles B1 and B2 , which is Von Neumann measurement under the condition of 1 X   0  1  . After measuring, Bob announces 2 his result. If Bob's result is  X   X , the B B 1

2

collapsed states are written as:

16


1 (6)  a ( x0 00  x1 01  x2 10  x3 11 )C1C2 4 Step3: After Alice and Bob announce their measurement results publicly, Charlie performs the joint unitary operation U2 on his particles C1C2 .

1 g8  ( 0001  0100  1011  1110 )a1a2 A1 A2 2

(9.8)

1 g9  ( 0010  0111  1000  1101 ) a1a2 A1 A2 2

(9.9)

1 g10  ( 0010  0111  1000  1101 ) a1a2 A1 A2 2

(9.10)

1 g11  ( 0010  0111  1000  1101 ) a1a2 A1 A2 2

(9.11)

1 g12  ( 0010  0111  1000  1101 ) a1a2 A1 A2 2

(9.12)

original state with a successful probability 4 a .

1 g13  ( 0011  0110  1001  1100 ) a1a2 A1 A2 2

(9.13)

The Second Scheme of the Two-Qubit State Telepertation

1 g14  ( 0011  0110  1001  1100 ) a1a2 A1 A2 2

(9.14)

1 g15  ( 0011  0110  1001  1100 ) a1a2 A1 A2 2

(9.15)

1 g16  ( 0011  0110  1001  1100 ) a1a2 A1 A2 2

(9.16)

C1C2

1  U2     

1 1

     1

(7)

the resulting state of Charlie’s particles will be the original state of |   a1a2 . Thus Charlie can get the 2

Arbitrary

The system state of particles is 

s

|   a1a2  C

A1 A2 B1 B2 C1C2

 [ x0 ( a 00000000  b 00010101  c 00101010  d 00111111 )  x1 (a 01000000  b 01010101

(8)

 c 01101010  d 01111111 )

Assuming that her measurement result is g1 , the composite system of particles B1 B2C1C2 becomes

 x2 (a 10000000  b 10010101

 c 10101010  d 10111111 )  x3 (a 11000000  b 11010101  c 11101010  d 11111111 )]a1a2 A1 A2 B1B2C1C2

We consider that Alice, Bob, and Charlie hold particles ( a1 , a2 A1 , A2 ),( B1 , B2 ) ( C1 , C2 ) respectively. Step1: Alice performs joint Bell measurement on her 4 qubits ( a1 , a2 A1 , A2 )respectively and then she announces her result publicly. The measurement basis is as follow:

17

1 g1  ( 0000  0101  1010  1111 )a1a2 A1 A2 2

(9.1)

1 g 2  ( 0000  0101  1010  1111 )a1a2 A1 A2 2

(9.2)

1 g3  ( 0000  0101  1010  1111 )a1a2 A1 A2 2

(9.3)

1 g 4  ( 0000  0101  1010  1111 )a1a2 A1 A2 2

(9.4)

1 g5  ( 0001  0100  1011  1110 ) a1a2 A1 A2 2

(9.5)

1 (ax0 0000  bx1 0101 2  cx2 1010  dx3 1111 ) B1B2C1C2

sub

(10)

Step2: Bob takes a joint X basis measurement Xˆ B1  Xˆ B2 on his two particles b1 and b2 , which is Von Neumann measurement under the condition of 1 X  ( 0  1 ) . After measuring, Bob announces 2 his result. If Bob's result is  X   X , the B1

B2

collapsed states can be written as: 1  C C  (ax0 00  bx1 01 1 2 4  cx2 10  dx3 11 )C1C2

(11)

Step3: Charile introduces an auxiliary two-state particle C with the initial state 0 to reincarnate C

C1C2

under the basis as follows: { 000

C1C2 C

, 010

C1C2 C

, 100

C1C2 C

, 110

C1C2 C

,

001 C C C , 011 C C C , 101 C C C , 111 C C C } , 1 2

1 g 6  ( 0001  0100  1011  1110 )a1a2 A1 A2 2

(9.6)

1 g 7  ( 0001  0100  1011  1110 )a1a2 A1 A2 2

(9.7)

1 2

1 2

1 (ax0 000  bx1 010 4  cx2 100  dx3 110 )C1C2C

C1C2 C

1 2

(12)

A collective unitary transformation U1 on particles

c1 , c2 , c may take the form of the following 8  8

JOURNAL OF OPTICS APPLICATIONS – Oct. 2012, Vol. 1, Iss. 2


matrix, namely

schemes, it is known that the first one is better than the

A U1   1  A2

A2   A1 

second one. In our first scheme, the receiver Charlie (13)

does not need introduce an auxiliary particle and operator unitary transformation. Only the sender Alice

Where A1 and A2 are 4  4 matrixs and may be

introduces

written

A1  diag m0 , m1 , m2 , m3 

measurement on her particles and the auxiliary

m0  1,

particle, the quantum state teleportation can be

as

and A2  diag{ 1  m02 , 1  m12 , 1  m22 , 1  m32 } if

m1  a , m 2  a , m 3 a , the unitary transformation b c d U1 will transform the state  C C C as follow: 1 2

C1C2 C

an

auxiliary

particle

and

makes

a

successfully realized with the maximal probability. The first scheme is more convenient for the receiver Charlie. Acknowledgements

1  (ax0 000  ax1 010 4  x1 b 2  a 2 011  ax2 100

(14)

 x2 c 2  a 2 101  ax3 110

This work is supported by the National Natural Science Foundation of China (Grant No. 10902083) and Shaanxi Natural Science Foundation under Contract

 x3 d 2  a 2 111 )C1C2C

(No. 2009JM1007).

Then Charile measures particle c on the basis of { 0 , 1 } . If his measured outcome is 1 , the quantum

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

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In summary, we have proposed two quantum state teleportation scheme of an arbitrary two-qubit state using a non-maximally entangled six-qubit Cluster state as channel. The receiver Charlie recovers the original state with a certain probability, the two scheme’s probability of success depends on the coefficient of the non-maximally entangled six-qubit 2

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Cluster state, the probability is  1 a   16  4  4 a 2 , 4  if a  b  c  d  1 , the probability will be 1. 2

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Qian Lan was born On November 27

Sinaca, vol.57(3)(in Chinese) , 2008.

1987 in Tianjin. she will get a

[15] Z. Yi , ‚ Experimental demonstration of time-shift attack against

practical

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controlled

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secure

direct

Greenberger–Horne–Zeilinger

master's degree in 2013. Now she is studying in

Xi’an University of

Posts

Telecommunications,

and

Xi’an, Shanxi China. Her major field of study is quantum communication. Email: lanqian198711@163.com

state‛.Optics Communications.vol.283, pp.192-195 ,2010.

19

JOURNAL OF OPTICS APPLICATIONS – Oct. 2012, Vol. 1, Iss. 2

Schemes of Teleportation with Six-qubit Cluster State  

In this paper, we have proposed a quantum state teleportation of an arbitrary two-qubit state using non-maximally entangled six-qubit Cluste...

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