量子隐形传态

Teleporting an unknown quantum state with unit
fidelity and unit probability via a non-maximally
entangled channel and an auxiliary system
Taghi Rashvand1,2
Received:5April2016/Accepted:21July2016/Published online:9August2016
©Springer Science+Business Media New York2016
Abstract We present a new scheme for quantum teleportation that one can teleport an unknown state via a non-maximally entangled channel with certainly,using an auxiliary system.In this scheme depending on the state of the auxiliary system,one canfind a class of orthogonal vectors set as a basis which by performing von Neumann measurement in each element of this class Alice can teleport an unknown state with unitfidelity and unit probability.A comparison of our scheme with some previous schemes is given and we will see that our scheme has advantages that the others do not.
Keywords Quantum teleportation·Non-maximally entangled channel
1Introduction
The transmission of information is one of the interesting processes of quantum infor-mation.Quantum teleportation is a process of transmission of information,a state of a system,from one place to another.For thefirst time in1993,Bennett et al.designed a scheme for teleportation[1].They used a system in a maximally entangled state as a quantum channel.In their scheme,Alice performs a Bell state measurement on her two systems,the system in an unknown state and one half of the maximally entangled system,and then sends her result to Bob who applies a local unitary operation to recover the unknown state.In this scheme,standard scheme,the existence of maximal entanglement is necessary.If one has a non-maximally entangled system,one has to B Taghi Rashvand
Taghi.rashvand88@
1Faculty of Science,University of Imam Khomeini International,Qazvin,Iran
2Tajhizsazanpishro Co.,Qazvin,Iran
4840T.Rashvand first convert the non-maximally entangled system to the maximally entangled system and then follow the standard scheme.Actually,in order to get fewer maximally entan-gled systems,it is necessary to provide many non-maximally entangled systems and perform a purification procedure[2]on them.
In the standard scheme,the existence of a Bell state measurement and a maximally entangled system are necessary.In other words,Alice and Bob have to share the max-imally entangled system and she also has to perform the Bell state measurement if they want to teleport correctly and completely.One can ask a question:Is there any scheme in which the existence of the maximally entangled system and the Bell state measurement are not necessary?To answer this question,many attempts have been carried out[3–11].Agrawal and Pati have proposed a scheme that uses orthogonal vectors as a basis and teleports,in general,an unknown state with unitfidelity but not unit probability,i.e.,teleportation is possible only for two out of four possible mea-surement results[6].Some others have proposed schemes in which Alice and/or Bob must do additional tasks,say,Alice has to perform a POVM instead of von Neumann measurement[3,4,9,11],prepare an auxiliary system[11],Bob has to prepare an aux-iliary system[5],and theyfirst concentrate entanglement of the channel[2].Although these schemes have advantages,they have a big problem:They are probabilistic.In other words,if one uses the previous schemes which use a non-maximally entangled system as a quantum channel and a von Neumann measurement,one cannot teleport an arbitrary unknown state with both unitfidelity and unit probability.Thus,it is one of the important advantages of our scheme that one can teleport an unknown state with both unitfidelity and unit probability.
In this article,we try to answer this question.We provide a new scheme using an auxiliary system in a specific state and perform a measurement in a specific basis. Actually in our scheme,Alice and Bob share a non-maximally entangled system,and Alice performs the measurement on three systems,auxiliary system,system in an unknown state and her entangled system,and then sends her results to Bob.Finally, Bob applies a local unitary operator to receive the unknown state.
We should mention that there are experimental works reported on successful tele-portation[12–14].
2Deterministic teleportation
In this section,we present a novel scheme to teleport quantum information determin-istically.In other words,we want to show how Alice can teleport an unknown state with unitfidelity and unit probability using our scheme.To do so,she has to prepare an auxiliary system and make a von Neumann measurement in a proper basis,which spans a eight-dimensional Hilbert space,on her systems.
Let us assume Alice and Bob share a non-maximally entangled system which is in the state
|ψ 23=L(|0 2|0 3+l|1 2|1 3),|L|2=1
,(1)
1+|l|2
where l is a known complex number except zero.The subscripts2and3label Alice and Bob’s system,respectively.Assume system1which Alice wants to teleport to Bob
Teleporting an unknown quantum state with unitfidelity (4841)
is in the state c1|0 1+c2|1 1and the auxiliary system is in the state a1|0 4+a2|1 4. The state of the whole system is
|ψ 1234=L(c1|0 1+c2|1 1)(|0 2|0 3+l|1 2|1 3)
×(a1|0 4+a2|1 4)
=Lc1a1|0 1|0 2|0 4|0 3+Lc1a2|0 1|0 2|1 4|0 3
+Llc1a1|0 1|1 2|0 4|1 3+Llc1a2|0 1|1 2|1 4|1 3
+Lc2a1|1 1|0 2|0 4|0 3+Lc2a2|1 1|0 2|1 4|0 3
+Llc2a1|1 1|1 2|0 4|1 3+Llc2a2|1 1|1 2|1 4|1 3.(2)
It is obvious that the dimension of Alice’s systems Hilbert space(as one system)is 8.So,to make a von Neumann measurement,she mustfind8orthonormal vectors which belong to this Hilbert space.If she follows our method[15],she mayfind that following8vectors,which are orthogonal to each other and have unit norm,and thus, form a von Neumann measurement,are proper(for proof see appendix):
|b1 =N11
La∗2
|001 +1
l∗
|111
,
|b2 =N2
1
a∗1
1
L
−ia
y
|000 +ia
ya∗2
|001
+N2
1
a∗1
1
l∗L
+ila
y
|110 −ila
a∗2y
|111
,
|b3 =N3
e
a∗1
|110 +1
a∗2
1
−l∗L
−e
|111
+N3
−(b+iy)
a∗1
|000 +1
a∗2
1
L
+b+iy
|001
,
|b4 =N4
1
a∗1
1
L
+c
|000 +
−c
a∗2
|001
+N4
1
a∗1
1
−l∗L
−lc
|110 +lc
a∗2
|111
,
|b5 =N51
La∗2
|101 +1
l∗
|011
,
|b6 =N6
1
a∗1
1
L
−ia
y
|100 +ia
ya∗2
|101
+N6
1
a∗1
1
l∗L
+ila
y
|010 −ila
a∗2y
|011
,
|b7 =N7
e
a∗1
|010 +1
a∗2
1
−l∗L
−e
|011
+N7
−(b+iy)
a∗1
|100 +1
a∗2
1
L
+b+iy
|101
,
4842T.Rashvand
|b8 =N8
1
a∗1
1
L
+c
|100 +
−c
a∗2
|101
+N8
1
a∗1
1
−l∗L
−lc
|010 +lc
a∗2
|011
,(3)
whereη=
a1a
2
2,b=1−η−|l|2−η|l|2
L(1+η)(1+|l|2)
,a=1−|l|2
|l|2(1+η)
,d=−2L
1+η
,y2=−(a+db)∈R,
e=l
L
1−1
|l|2
+l(b+iy),c=d−iy,and the parameters N i,i=1,2,...,8
can be used to normalize the vectors|b i ,i=1,2,...,8,respectively.We should mention that in the above equation,for convenience,the term|... is used instead of | 1| 2| 4.
It is not difficult to show that the following8parameters can normalize vectors |b i ,i=1,2,...,8(L∈R):
1 |N1|2=1
|N5|2
=1
L2|a2|2
1+
1
|l|2
,
1 |N2|2=1
|N6|2
=1
L2|a1|2
a2
y2
(1+η)+1
L2|l|2
,
1 |N3|2=1
|N7|2
=1
|a1|2
1
L2
1
|l|2
−3
−4b
L
+1
|a1|2
(1+η)
1+|l|2
1
L
+b
2
+y2
,
1 |N4|2=1
|N8|2
=1
|a1|2
1
L4|l|2
+4d
L
+1
|a1|2
(1+η)
1+|l|2
d2+y2
.(4)
As we see,infinite bases which may be used in our scheme to teleport the unknown state with unitfidelity and unit probability were presented.Also,these bases are related to the state of the auxiliary system and they exist for any choice of a1and a2belonging to the set of complex numbers except zero,i.e.,a1,a2∈C−{0},and obeying|a1|2+|a2|2=ing each of these bases,we can express each| 1| 2| 4
in terms of its vectors,and thus we can rewrite Eq.(2)as
|ψ 1234=N∗1(c1|0 3+c2|1 3)|b1
+N∗2(c1|0 3+c2|1 3)|b2
+N∗3(c1|0 3−c2|1 3)|b3
+N∗4(c1|0 3−c2|1 3)|b4
+N∗5(c2|0 3+c1|1 3)|b5
+N∗6(c2|0 3+c1|1 3)|b6
Teleporting an unknown quantum state with unitfidelity (4843)
+N∗7(c2|0 3−c1|1 3)|b7
+N∗8(c2|0 3−c1|1 3)|b8 .(5) It is clear from Eq.(5)that after Alice performs a measurement in the basis{|b i }8i=1, Bob’s system will collapse into one of the four following equations:
c1|0 3±c2|1 3,
c2|0 3±c1|1 3.(6)
Thus,according to Alice’s measurement outcome,Bob must apply one of the four local operators
ˆ1,
10
0−1
,
01 10
,
0−1
10
,
(7)
in order to convert his system state to c1|0 3+c2|1 3.
We have shown how one can use this scheme to teleport an unknown state with unitfidelity and unit probability using an auxiliary system together non-maximally entangled system as quantum channel.
Let us conclude this section with a discussion of the entanglement measure and efficiency of our scheme.There are several measures of entanglement,but the entropy of entanglement can be accepted as unique measure of entanglement for pure states [2].It is defined as S(ρ)=−tr{ρlogρ2},whereρis density operator of system 2or system3.One may obtain the entanglement measure of Eq.(1)is given by
S(ρ)=−L2log L22−L2|l|2log L2|l|2
2[6,16].Since l is any nonzero complex number,
it is not difficult to see that0<S(ρ)≤1.This means that entanglement measure of quantum channel,Eq.(1),is a function of l and belongs to(0,1](we note that S(ρ)=0,which is corresponding to separate state,i.e.,has no entanglement,does not include).Thus,it is not important that the quantum channel has what amount of entanglement and Alice can teleport an unknown state with unitfidelity and unit probability.In other words,to teleport an unknown state with both unitfidelity and unit probability,it is only need that quantum channel to be entanglement,independent of its amount.
3Compare our scheme with some other schemes
Because of the many advantages of non-maximally entangled states[5],much atten-tion has been paid to the non-maximally entangled states and many schemes were proposed to teleport an unknown quantum state using non-maximally entangled state as a quantum channel[3–11].Since these schemes cannot maintain unitfidelity and unit probability simultaneously,one can categorize them in two kinds:ones that main-tain unit probability and leave unitfidelity,and ones that maintain unitfidelity but not
4844T.Rashvand unit probability.If one teleports a state and want her(his)state is exactly reproduced on the receiver’s side,thefirst ones are not good.So,let us discuss the second ones.In some sense,later can be divided into four kinds:first,ones in which an auxiliary system is used;second,ones in which generalized measurements,instead of von Neumann measurement,are used;third,ones in which both an auxiliary system and generalized measurements are used;and fourth,ones carried out like standard teleportation,i.e., a von Neumann measurement is used and no auxiliary system is necessary.All of these schemes have a common feature that may be a big problem:These schemes are probabilistic.
Let us look at some of these schemes in detail and then compare them with our scheme.
3.1Scheme of Agrawal and Pati
In2002,Agrawal and Pati proposed a quantum teleportation scheme using non-maximally entangled state and von Neumann measurement.Their scheme belongs to fourth kind,i.e.,to teleport an unknown quantum state using non-maximally entan-gled system as quantum channel,Alice performs a von Neumann measurement and Bob performs a proper unity operation.
As they said[6,8],the difference between their scheme and the existing ones is that they use neither localfiltering nor entanglement concentration.Their scheme works just with two classical bits.In our scheme,also,we never need to use local filtering or entanglement concentration and our scheme works with two classical bits (In order to send her result to Bob,Alice has two options(at least):(1)she sends classical massage that contains information about the vector which is her result.To do so,she needs3classical bits.(2)she sends classical massage contains information about the local operator that Bob must perform to receive the initial state.As we see in Eq.(7),there are four local operations.Thus,she needs just two classical bits).In our scheme like their scheme Bob performs just a proper unity operation and Alice makes a von Neumann measurement.So,in some aspects our scheme is similar to their scheme.Although our scheme in some aspects is similar to their scheme,they have very important differences.Ours is deterministic,but theirs is probabilistic.In our scheme,Alice can perform measurement in any proper basis that she wants,but in their scheme she cannot.
We would like to point out that the scheme of Agrawal and Pati is improved via two different methods(i.e.,the auxiliary system and generalized measurement)by Xu and others[9].They increase its success probability,but not to unit,i.e.,their schemes are still probabilistic.
3.2Scheme of Li and coworkers
Li et al.have proposed a scheme to transmit an unknown state using non-maximally entangled system[5].In their scheme,Alice performs a von Neumann measurement on her system and Bob mustfirst prepare an auxiliary system in the state|0 ,then perform a collective unitary transformation on his two systems andfinally perform
Teleporting an unknown quantum state with unitfidelity (4845)
a measurement on auxiliary system.If the result of his measurement is|0 ,the tele-portation is successfully accessed,while if his result is|1 ,teleportation fails.With their scheme,one can teleport an unknown state with unitfidelity,but not with unit probability.
Our scheme,in some aspects,is similar to the scheme of Li and coworkers although they are extremely ing an auxiliary system in both schemes is necessary. In our scheme,Alice must prepare the auxiliary system,but in the scheme of Li and coworkers Bob prepares it.In both schemes,a von Neumann measurement is used.In the scheme of Li and coworkers,Alice can perform a measurement in non-maximally entangled eigenstates and Bob must also perform a measurement while in our scheme only Alice must perform a measurement on his three systems.In other words,in our scheme,after Alice performs a measurement on his system,the state of Bob’s system will collapse to one of the Eqs.(6),i.e.,to reestablish the initial state,Bob just needs to perform a proper unitary transformation.In the scheme of Li and coworkers,Alice has to perform a measurement in a special basis while in our scheme she can choose one of the infinite bases.
3.3Scheme of Bandyopadhyay
Bandyopadhyay proposed two methods of teleporting an unknown state using a non-maximally entangled channel.In hisfirst method,Alice must prepare an auxiliary system beside a system that he wants to teleport the state of it to Bob.The state parameter of auxiliary system must be the Schmidt coefficients of the non-maximally entangled channel.Alice now performs a measurement on her three particles in a special basis.Depending on the outcome of Alice’s measurement,she may also need to perform a POVM measurement.So she must send three classical bits to Bob if she gets necessary to perform a POVM measurement.Finally,according to information that Alice sent to Bob,he must perform a proper unitary transformation to reconstruct the unknown state.
In this scheme of Bandyopadhyay like our scheme,Alice must perform a measure-ment on three particles and Bob only perform a unitary transformation.In his scheme, she must perform a measurement in special basis,but in our scheme she has choices. In his scheme,Alice sometime needs to perform POVM measurement and sends three classical bits while in our scheme she performs only von Neumann measurement and sends two classical bits.
In his second scheme,Alice performs a Bell state measurement on her two systems (her system in entangled state and system in an unknown state)and sends her result to Bob.This step of his scheme is like one of the standard teleportation schemes.But Bob cannot reconstruct the initial state if he follows standard teleportation.To reconstruct the initial state,Bob must prepare an auxiliary system in a state like Li and coworkers’scheme,perform a CNOT operation on his two systems and then perform a POVM measurement.
In both schemes of Bandyopadhyay,either sender(Alice)or receiver(Bob)per-forms a POVM and the unknown state could not be deterministically teleported to Bob,while in our scheme,Alice performs a von Neumann measurement and quantum teleportation is deterministic.
4846T.Rashvand In summary,Mor and Horodecki[3]have tried to propose a scheme of teleportation in which one does not need to use maximally entangled channel.The cost that they pay to do this is generalized measurement and conclusive teleportation[3].Bandyopad-hyay has tried to propose a better scheme that the success probability of teleportation is optimal[4].The advantage of the scheme of Li and coworkers is that von Neumann measurement,instead of POVM measurement,is used[5].Others[6–11]have also tried to propose a scheme that has advantages that the others do not have.However,all of these schemes are probabilistic,i.e.,an unknown quantum state can be teleported with unitfidelity,but not unit probability.It is the advantage of our scheme that use von Neumann measurement and the unknown quantum state can be teleported with both unitfidelity and unit probability.
We would like to point out that some ones have proposed schemes to teleport an unknown entangled state of two systems via an entangled three-system state as a quantum channel[17,18].These schemes are also probabilistic.To solve this problem, one can use our method.
4Conclusion and summary
In summary,a scheme to teleport an unknown quantum state with unitfidelity and unit probability using a non-maximally entangled channel is presented.The cost that one has to pay to achieve this goal is to perform a measurement in an eight-dimensional Hilbert space which is twice larger than the conventional Hilbert space which is made by four Bell states.Although Alice performs a measurement in a basis which has8 elements,the number of classical bits that she needs to send to Bob is2.Our scheme was compared with other schemes and was shown that our scheme has a major advantage over them.First,in our scheme teleportation is carried out with both unitfidelity and unit probability while,in other schemes is not.Some schemes maintain unitfidelity, but have to leave unit probability and some others maintain unit probability,but leave unitfidelity.As well known,maintaining both unitfidelity and unit probability is very important in quantum information processing.Second,in our scheme,Alice can choose one of the infinite bases to perform a measurement according her laboratory instruments.With each basis that is chosen,teleportation scheme is deterministic. Third,in our scheme,Alice performs only a von Neumann measurement and Bob carries out one proper unitary transformation that depends on the result of Alice’s measurement.
Appendix
In this appendix,we want to prove that Eq.(3)makes von Neumann measurement.In other words,the set of vectors|b i ,i=1,2,...,8,i.e.,
ξ={|b1 ,|b2 ,|b3 ,|b4 ,|b5 ,|b6 ,|b7 ,|b8 }
is an orthonormal basis for the Hilbert space of Alice’s systems(systems1,2and 4)[19–21].Note that since the Hilbert space of Alice’s systems is eight-dimensional
Teleporting an unknown quantum state with unit fidelity...4847space,an orthonormal basis is a set of 8vectors which are orthogonal to each other and the norm of them is one,that is b i |b j = 0i =j 1i =j
.As we see,if we choose N i ,i =1,2,...,8as Eq.(4),then the norm of each vector in the set ξis one.So,to prove that Eq.(3)makes a von Neumann measurement,it is efficient to prove all vectors in the set ξare orthogonal to each other.
Let us start by dividing the set ξinto two sets ξ1={|b 1 ,|b 2 ,|b 3 ,|b 4 }and ξ2={|b 5 ,|b 6 ,|b 7 ,|b 8 }.It is not difficult to see that each vector in the set ξ1is orthogonal to each vector in the set ξ2.So,we only need to show that every distinct pair in the set ξ1and in the set ξ2is orthogonal.In what follows,we just prove that all vectors in ξ1are orthogonal to each other,because one can do same calculation for ξ2.First,we show |b 1 is orthogonal to three others:
b 1|b 4 =N ∗1N 4
c L |a 2|2+N ∗1N 4−c L |a 2|2
=0, b 1|b 2 =N ∗1N 2ia Ly |a 2|2+N ∗1N 2−ia Ly |a 2|2=0, b 1|b 3 =N ∗1N 31L |a 2| 1L +b +iy +N ∗1N 31L |a 2|21l 1−l ∗L −e =N ∗1N 31L |a 2|2 1L 1−1|l |2 +b +iy −e l .(8)
As mentioned in Sect.2that e l =1L
1−1|l |2 +b +iy ,thus b 1|b 3 =0.Now,we try to prove b 2|b 3 =0.It is easy to show that b 2|b 3 |a 1|2N ∗2N 3= 1L +i a y (−b −iy )+η−ia y 1L +b +iy + 1L −i |l |2a y e l +ηia −ηil ∗a y e .(9)
Let us first show that the real part of Eq.(9)is zero.By using parameters which were defined in Sect.2,one can simply derive the real part of Eq.(9)is as
Re b 2|b 3 |a 1|2N ∗2N 3 =1L 2 |l |2−1|l |2
+a (1+η) 1+|l |2 .(10)
4848T.Rashvand
By substituting1
L2=
1+|l|2
and a=1−|l|2
|l|2(1+η)
,it follows that the real part of Eq.(9)
is zero.We now turn our attention to the calculation of the imaginary part of Eq.(9)
I m
b2|b3
|a1|2
N∗2N3
=
−ba
y
(1+η)
1+|l|2
+a
Ly
1−η−|l|2−η|l|2
.(11)
Since b(1+η)
1+|l|2
L=1−η−|l|2−η|l|2,it follows that the imaginary part
of Eq.(9)is zero.
In order to see|b2 and|b4 are orthogonal,one can obtain
b2|b4 |a1|2
N∗2N4
=
1
L
+i a
y
1
L
+c
+ηia
y
(c)+
1
l L
−i l
∗a
y
1
−l∗L
−lc
+ηil
∗a
y
(lc).(12)
By substituting c=d−iy,one may rewrite Eq.(12)in the form
b2|b4 |a1|2
N∗2N4
=
1
L
+i a
y
1
L
+d−iy
+ηia
y
(d−iy)
1+|l|2
+
1
l L
−i l
∗a
y
1
−l∗L
−ld+liy
.(13)
It is not difficult to see that the real part of Eq.(13)is
Re
b2|b4
|a1|2
N∗2N4
=1
L2
1−
1
|l|2
+a(1+η)
1+|l|2
.(14)
As Eq.(10),the right-hand side of Eq.(14)is zero.One can simplyfind that the imaginary part of Eq.(13)as
I m
b2|b4
|a1|2
N∗2N4
=2
L
a
y
+a
y
d
1+η+|l|2+η|l|2
.(15)
By substituting1
L2=
1+|l|2
and d=−2L
1+η
,it follows that the imaginary part of
Eq.(13)is zero.Thus, b2|b4 =0.
Teleporting an unknown quantum state with unitfidelity (4849)
Finally,we try to show that|b3 and|b4 are orthogonal
b2|b4 |a1|2
N∗2N4
=−e
l L
−e
l
|l|2c∗−
ηc∗
L
−ηe
l
|l|2c∗−(b+iy)1
L
−(b+iy)c∗
−η1
L
c∗−(b+iy)ηc∗.(16)
By defining t to be
e l =1
L
1−
1
|l|2
+(b+iy) =t+(b+iy),(17)
one can rewrite Eq.(16)as
b2|b4 |a1|2
N∗2N4
=−t
L
−|l|2c∗(1+η)t
−2
L
(b+iy)−(b+iy)c∗(1+η)1
L2
−2ηc
∗
L
.(18)
The real part of Eq.(18)is
Re
b2|b4
|a1|2
N∗2N4
=−t
L
−|l|2(1+η)td
−2
L
b−(1+η)
1
L2
bd−y2
−2ηd
L
.(19)
Let us simplify some terms of Eq.(19)separately,using parameters which were defined in Sect.2
(1+η)1
L2
bd−y2
=1
L2
(1+η)(2bd+a)=
=−4b
L
+1
L2
1
|l|2
−1
=4b
L
−t
L
,
|l|2(1+η)td=−2|l|2Lt=−2|l|2+2,
2
L
b=
1−η−|l|2−η|l|2
L(1+η)
1+|l|2
2
L
=21−η
1+η
−2|l|2, 2ηd
L
=
−4η
1+η
.(20)
Thus,after combining Eq.(19)and Eq.(20),it is not difficult to show that the real part of Eq.(18)is zero.The imaginary part of Eq.(18)may be written as
123
4850T.Rashvand
I m
b2|b4
|a1|2
N∗2N4
=−|l|2(1+η)ty−2
L
y
−(by+yd)(1+η)1
L2+2ηy
−L
.(21)
Since
d(1+η)1
L2
=
−2
L
,
b(1+η)1
L2
=
−|l|2
L
(1+η)+1−η
L
,
|l|2(1+η)t=|l|2(1+η)1
L −(1+η)1
L
,(22)
we clearly see that the imaginary part of Eq.(18)is zero.So, b3|b4 =0. References
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