Simultaneous quantum state teleportation via the locked entanglement channel

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by  Mei-Yu Wang

Simultaneous quantum state teleportation via the locked entanglement channel
Meiyu Wang a, Fengli Yan a, Ting Gao b, and Youcheng Li a
∗
a College of Physics and Information Engineering,
Hebei Normal University, Shijiazhuang 050016, China
b College of Mathematics and Information Science,
Hebei Normal University, Shijiazhuang 050016, China
(Dated: February 1, 2008)
Wepresentasimultaneousquantumstateteleportationscheme,inwhichreceiverscannotrecover
their quantum state separately. When they want to recover their respective quantum state, they
must perform an unlocking operator together.
PACSnumbers: 03.67.Hk
8
0
Quantumentanglementplaysanimportantroleinvar- teleportedtothemrespectivelybylockingtheirquantum
0
ious fields of quantum information, such as quantum channels.
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computation, quantum cryptography, quantum telepor- To present our scheme clearly, let’s first begin with
n
tattion and dense coding etc. Quantum teleportation is simultaneous quantum state teleportation between one
a
J one of the most important applications of quantum en- sender and two receivers.
6 tanglement. In quantum teleportation process, an un- The quantum states of two qubits T1 and T2 to be
2 known state can be transmitted from a sender Alice to teleported are as follows
a receiver Bob without transmission of carrier of quan-
2 tum state. Since Bennett et al [1] presented a quan- |φ1iT1 =α1|0iT1 +β1|1iT1, (1)
v tum teleportation scheme, there has been great develop- |φ2iT2 =α2|0iT2 +β2|1iT2,
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mentintheoreticalandexperimentalstudies. Nowquan-
1 tum teleportation has been generalized to many cases Alice wants to teleport φ1 to Bob and φ2 to Char-
0 lie simultaneously. Suppo|seithat Alice, Bob| aind Charlie
[2, 3, 4, 5, 6, 7, 8, 9]. Moreover, quantum teleportation
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share two EPR pairs denoted as
1 hasbeendemonstratedwiththepolarizationphoton[10]
06 amnedntas.siTnhgeletceolehpeorretnacteiomnoodfeaocfohfieerldenst[1st1a]tienctohrereesxppoenrdi-- |EPRi1 = √12(|00iA1B +|11iA1B), (2)
h/ ingtocontinuousvariablesystemwasalsorealizedinthe |EPRi2 = √12(|00iA2C +|11iA2C),
p laboratory [12].
- In the original proposal, the teleportation of a single where A1 and A2 belong to Alice, B and C belong to
t Bob and Charlie respectively. Then the quantum state
n qubit φ = a0 +b1 is executed as follows: Alice and
a | i | i | i ofthejointsystem(qubitstobeteleportedandtwoEPR
Bob initially share an EPR pair as a quantum channel.
u pairs) can be written as
Aliceperformsajointmeasurementonthecomposedsys-
q
Xiv: tpseiacmiarl).c(qhSuahnbenitterlt.aonBsbomebitatspeltpehlpieeosortuaetcdcooramrneedsptooonnBedoinbogftuhthnroeituaegrnhytaaonpcgellareasd-- |Φi=(⊗α121α(2|0|0000i0+i+α|10β120|011ii++|1α021β01i|1+0i|1+11β11iβ)A2|11A12iB)TC1.T2
(3)
tiononhisparticleoftheentangledpair,whichischosen
r The scheme of simultaneous teleportation consists of
a in accordance with the outcome of joint measurement.
the following five steps.
ThefinalstateofBob’squbitiscompletelyequivalentto
(S1) Locking the quantum channels
the original unknown state.
In this step of teleportation, Bob and Charlie perform
If Alice wants to teleport the quantum state φ1 to a joint unitary transformation
| i
Bob and teleport φ2 to Charlie, obviously two EPR
| i
pairs are required. One is shared between Alice and 1 0 1 0
Bob; the other is shared between Alice and Charlie. 1 0 1 0 1
U =   (4)
The quantum state teleportation can be completed by BC √2 0 1 0 1
Bennett’s protocol. But if Bob and Charlie want to re- 1 0 1 −0 
 − 
ceivetheirrespectivequantumstatesimultaneously,how
do they complete the teleportation? In this paper, we on the particles B and C. After that, the state of the
will present a simultaneous quantum state teleportation total system becomes
scheme. In this scheme, the recievers Bob and Charlie
Φ
cansynchronouslyrecoverthequantumstatewhichAlice | ′i
=(α1α12|0(00i0+00α1+β20|0011i1++α20β110|110i++0β111β02|11i)T1T2
⊗ 2√2 | i | i | i | i
+ 1000 1011 + 1101 1110 ) .
| i−| i | i−| i A1A2BC
∗Correspondingauthor. Emailaddress: fl[email protected] (5)
2
(S2) Performing Bell-basis measurement the Peres-Horodeckicriterion[13,14]fortwoqubits that
Alice performsaprojectivemeasurementonA1T1 and theirdensityoperatorρisseparableifandonlyifitspar-
one on A2T2 in the Bell-basis Ψm ,m = 0,1,2,3 , tialtranspositionispositive. Thepartialtranspositionof
{| i }
where ρ is defined as
|ΨΨ02i== √112((|0000i+|1111i)),, |ΨΨ31i== √112((|0011i+|1100i))., (6) ρT =Σijklρjikl|iihj|⊗|kihl|, (12)
| i √2 | i−| i | i √2 | i−| i where ρ = Σ ρ i j k l . After transformation,
ijkl ijkl
| ih |⊗| ih |
Let σ be one member of the set of rotation matrices the reduced density operator of (B,C) is given as
j
1
0 1 1 0 0 1 ρ = I. (13)
I, , , (7) ′BC 4
{ (cid:18)1 0(cid:19) (cid:18)0 1(cid:19) (cid:18) 1 0 (cid:19)}
− −
The partialtranspositionofρ has onlypositive eigen-
composed of the identity operator and three Pauli spin values 1,1,1,1. The result i′BmCplies that there is no en-
operators. ItiseasytoprovethatEq.(5)canberewritten 4 4 4 4
tanglementbetweenBandC. Infact,beforetransforma-
as tionthe reduceddensity operatorof(B,C) is ρ = 1I.
BC 4
For any unitary operator U we always have
|Φ′i=Σm,n|ΨmiA1T1|ΨniA2T2UBCσm|φ1iB⊗σn|φ2iC.
(8) 1
After the projective measurement, the state of particles ρ′BC =U+ρBCU =ρBC = 4I, (14)
B and C collapses into
i.e. there is no entanglement between B and C for arbi-
Ψ BC =UBCσm φ1 B σn φ2 C, (9) trary unitary operator U. Notice that after transforma-
| i | i ⊗ | i
tion, the quantum channels become
whichcorrespondstothemeasurementresult Ψ , Ψ .
m n
(S3) Transmitting the measurement outcom|e i | i Ψ = 1 (0000 + 0011 + 0101 + 0110
| ′i 2√2 | i | i | i | i
After performing the Bell-basis measurement, Alice + 1000 1011 + 1101 1110 )
transmits the outcome of measurement (i.e. m and n) = 1 [(|00 +i−10| ) i |(00 i+−1|1 i)A1A2BC
to Bob and Charlie. 2√2 | i | i A1B ⊗ | i | iA2C
+(01 11 ) (01 + 10 )].
(S4) Unlocking the quantum state | i−| i A1B ⊗ | i | iA2C
(15)
According to Eq.(9) the quantum state of particle B
Eq. (15)shows that Ψ is maximallyentangledstate of
andC islocked,soBobandCharliemust”unlock”Band | ′i
C firstly. In order to recover the quantum states which A1B and A2C. In other words, the function of UBC is
not to entangle (B,C) but (A1B,A2C). In some sense,
are teleported from Alice, Bob and Charlie perform a
U islikea”lock”whichpreventsBobandCharliefrom
unitary operator U+ on the particles B and C, where BC
BC recoveringtheirquantumstatesseparately. Moresurpris-
1 0 0 1 ingly, the initial state of (A1,B) is maximally entangled,
while after performing the unitary transformation U
1 0 1 1 0 BC
U+ =  . (10) the reduced density operator ρ reads
BC √2 1 0 0 1 ′A1B
−
 0 1 1 0 
 −  1 0 1 0
1 0 1 0 1
Ainfttoer that the state of particle B and C is transformed ρ′A1B = 41 0 1 −0 . (16)
0 1 0 1 
 − 
Ψ
′ BC
=|UB+iCUBCσm|φ1iB ⊗σn|φ2iC =σm|φ1iB⊗σn|φ2i(C11.) Teihgeenpvaarlutieasl0t,r0a,n1sp,o1s.itTiohneorfesρu′Alt1Bimhpalsieosntlhyatnotnhneeegnattaivne-
2 2
(S5) Recovering the quantum state glement of (A1,B) has vanished. Similarly, the entan-
Bob and Charlie perform a local unitary operator σm glement of (A2,C) has also vanished. That is, A1B and
and σn on particles B and C respectively, then they will A2C have no function of the quantum channels, after
obtain φ1 and φ2 immediately. UBC is applied on B and C particles. Alice can teleport
| i | i
In the following, we will discuss why we call the step φ1 and φ2 to Bob and Charlie respectively due to the
| i | i
(S1) ”locking” the quantum channel. It is easy to ver- entanglement between A1B and A2C.
ify that U can be realized by a Hadamard operator It is worthy pointing out that ”locking” the quantum
BC
on particle B and a CNOT operator (B is control qubit channels can be completed by Alice. On the one hand,
and C is target qubit). Apparently U is a non-local beforedistributingtheentanglementpairs,Alicedoesthe
BC
operator. Now, is there any entanglement between B transformationU onthetwoqubitswhichwillbesent
BC
and C after Bob and Charlie make the unitary opera- to Bob and Charlie, i.e. on qubits B and C. The advan-
tor U on them? The answer is negative. To inves- tage of which is that Bob and Charlie need not come
BC
tigate the entanglement between B and C, we employ together to lock the states. They only need to come
3
together at the time they want to unlock it. On the It is only necessary to replace locking operator U
B1B2
otherhand,AlicemayalsomakeunitaryoperatorU and unlocking operator U+ by unitary operator
A1A2 B1B2
onqubits A1 andA2 after distributing the entanglement
pairs UB1B2 Bn =Πni=2(CNOT)B1BiHB1 (20)
···
U I 1(0000 + 0101 + 1010 + 1111 ) andU+ respectively. Here H is a Hadmardop-
= A11A(20⊗000BC+2 0|011 i+ 0|101 i+ 0|110 i+ 1|000 i A1A2BC eratorBo1nB2t·h··eBnqubit B1. B1
2√2 | i | i | i | i | i
1011 + 1101 1110 ) , Without difficulty one can figure out the network for
−| i | i−| i A1A2BC (17) locking the n quantum channels, for saving space we do
where not depict it here.
Insummary,wepresentasimultaneousquantumstate
1 0 0 1 teleportation scheme, in which receivers can not recover
U = 1  0 1 1 0 . (18) their quantum state separately. When they want to re-
A1A2 √2 1 0 0 1 covertheirquantumstates,they mustperformaunlock-
 0 1 1 −0  ing operator together.
 − 
It is straightforward to generalize the above protocol
to n receivers who recover the quantum states simulta-
Acknowledgments
neously. Suppose Alice wants to teleport φ to Bob i
i
| i
(i = 1,2,...,n), Alice and every receiver share an EPR
This workwassupportedbythe NationalNaturalSci-
pair denoted as
ence Foundation of China under Grant No: 10671054,
1 HebeiNaturalScience FoundationofChina under Grant
EPR = (00 + 11 ),i=1,2,...,n. (19)
| ii √2 | iAiBi | iAiBi Nos: A2005000140,07M006 and the Key Project of Sci-
ence and Technology Research of Education Ministry of
Thestepsinthiscasearesimilartothatoftworeceivers. China under Grant No:207011.
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