Quantum Teleportation and Beam Splitting

Transcription

1 Quantum Teleportation and Beam Splitting arxiv:quant-ph/ v1 11 May 000 Karl-Heinz Fichtner Friedrich-Schiller-Universität Jena Fakultät für Mathematik und Informatik Institut für Angewandte Mathematik D Jena Deutschland and Masanori Ohya Department of Information Sciences Science University of Tokyo Chiba Japan Abstract Following the previous paper in which quantum teleportation is rigorously discussed with coherent entangled states given by beam splittings, we further discuss two types of models, perfect teleportation model and non-perfect teleportation model, in general scheme. Then the difference among several models, i.e., the perfect models and the non-perfect models, is studied. Our teleportation models are constructed by means of coherent states in some Fock space with counting measures, so that our model can be treated in the frame of usual optical communication. 1 Introduction Following the previous paper [1], we further discuss the non-perfect teleportation. The notion of non-perfect teleportation is introduced in [1] to 1

2 construct a handy i.e., physically more realizable) teleportation, although its mathematics becomes a little more complicated. For the completeness of the present paper, we quickly review the meaning of the teleportation and some basic facts of Fock space in this section. Then we dicuss the perfect teleportation in very general more general than one given in [1]) scheme with our previous results, and we state the main theorem obtained in [1] for non-perfect teleportation, both in the section. The main results of this paper are presented in the section 3, where we discuss the difference among three models, i.e., the perfect model, the non-perfect one given in [1] and that discussed in the present paper. The proofs of the main results are given in the section Quantum teleportation The study of quantum teleportation was started by the paper [3] as a part of quantum cryptography [5], whose scheme can be mathematically expressed in the following steps [11,, 1]: Step 0: A girl named Alice has an unknown quantum state ρ on a dimensional) Hilbert space H 1 and she was asked to teleport it to a boy named Bob. Step 1: For this purpose, we need two other Hilbert spaces H and H 3, H is attached to Alice and H 3 is attached to Bob. Prearrange a so-called entangled state σ on H H 3 having certain correlations and prepare an ensemble of the combined system in the state ρ σ on H 1 H H 3. Step : One then fixes a family of mutually orthogonal projections F nm ) n,m=1 on the Hilbert space H 1 H corresponding to an observable F := z n,m F nm, and for a fixed one pair of indices n, m, Alice performs a n,m first kind incomplete measurement, involving only the H 1 H part of the system in the state ρ σ, which filters the value z nm, that is, after measurement on the given ensemble ρ σ of identically prepared systems, only those where F shows the value z nm are allowed to pass. According to the von eumann rule, after Alice s measurement, the state becomes ρ 13) nm := F nm 1)ρ σf nm 1) tr 13 F nm 1)ρ σf nm 1)

3 where tr 13 is the full trace on the Hilbert space H 1 H H 3. Step 3: Bob is informed which measurement was done by Alice. This is equivalent to transmit the information that the eigenvalue z nm was detected. This information is transmitted from Alice to Bob without disturbance and by means of classical tools. Step 4: Making only partial measurements on the third part on the system in the state ρ 13) nm means that Bob will control a state Λ nm ρ) on H 3 given by the partial trace on H 1 H of the state ρ 13) nm after Alice s measurement) Λ nm ρ) = tr 1 ρ 13) nm = tr 1 F nm 1)ρ σf nm 1) tr 13 F nm 1)ρ σf nm 1) Thus the whole teleportation scheme given by the family F nm ) and the entangled state σ can be characterized by the family Λ nm ) of channels from the set of states on H 1 into the set of states on H 3 and the family p nm ) given by p nm ρ) := tr 13 F nm 1)ρ σf nm 1) of the probabilities that Alice s measurement according to the observable F will show the value z nm. The teleportation scheme works perfectly with respect to a certain class S of states ρ on H 1 if the following conditions are fulfilled. E1) For each n, m there exists a unitary operator v nm : H 1 H 3 such that Λ nm ρ) = v nm ρ v nm ρ S) E) p nm ρ) = 1 ρ S) nm E1) means that Bob can reconstruct the original state ρ by unitary keys {v nm } provided to him. 3

4 E) means that Bob will succeed to find a proper key with certainty. Such a teleportation process can be classified into two cases [], i.e., weak teleportation and general teleportation, in which the solutions of the teleportation in each case and the conditions of the uniqueness of unitary key were discussed. The solution of the weak teleportation is a triple { σ 3), F 1), U } such that Λ ρ 1) = U ρ 1) U holds for any state ρ 1) SH 1 ). Once a weak solution of a teleportation problem is given, we can construct the general solution for all n, m above []. In [1], we considered a teleportation model where the entangled state σ is given by the splitting of a superposition of certain coherent states, although this model doesn t work perfectly, that is, neither E) nor E1) hold. In the same paper, we estimated the difference between the perfect teleportation and this non-perfect teleportation by adding a further step in the teleportation scheme: Step 5: Bob will perform a measurement on his part of the system according to the projection F + := 1 exp0) >< exp0) where exp0) >< exp0) denotes the vacuum state the coherent state with density 0). Then our new teleportation channels we denote it again by Λ nm ) have the form Λ nm ρ) := tr 1 F nm F + )ρ σf nm F + ) tr 13 F nm F + )ρ σf nm F + ) and the corresponding probabilities are p nm ρ) := tr 13 F nm F + ) ρ σf nm F + ) For this teleportation scheme, E1) is fulfilled but E) is not, about which we review in the next section. 4

5 1. Basic otions and otations We collect some basic facts concerning the symmetric) Fock space in a way adapted to the language of counting measures. For details we refer to [6, 7, 8,, 9]. Let G be an arbitrary complete separable metric space. Further, let µ be a locally finite diffuse measure on G, i.e. µb) < + for bounded measurable subsets of G and µ{x}) = 0 for all singletons x G. We denote the set of all finite counting measures on G by M = MG). Since ϕ M can be written in the form ϕ = n δ xj for some n = 0, 1,,... and x j G with the Dirac measure δ x corresponding to x G, the elements of M can be interpreted as finite symmetric) point configurations in G. We equip M with its canonical σ algebra W cf. [6], [7]) and we consider the σ finite measure F by setting FY ) := X Y O) + n ) 1 δ xj µ n d[x 1,..., x n ])Y W), n! n 1 j=1 G n X Y where X Y denotes the indicator function of a set Y and O represents the empty configuration, i. e., OG) = 0. Since µ was assumed to be diffuse one easily checks that F is concentrated on the set of a simple configurations i.e., without multiple points) G. j=1 ˆM := {ϕ M ϕ{x}) 1 for all x G} M = MG) := L M, W, F) is called the symmetric) Fock space over In [6] it was proved that M and the Boson Fock space ΓL G)) in the usual definition are isomorphic. For each Φ M with Φ 0 we denote by Φ > the corresponding normalized vector Φ >:= Φ Φ. Further, Φ >< Φ denotes the corresponding one dimensional projection describing a pure state given by the normalized vector Φ >. ow, for each 5

6 n 1 let M n be the n fold tensor product of the Hilbert space M, which can be identified with L M n, F n ). For a given function g : G C the function exp g) : M C defined by { 1 if ϕ = 0 exp g) ϕ) := x G,ϕ{x})>0 gx) otherwise is called exponential vector generated by g. Observe that exp g) M if and only if g L G) and one has in this case exp g) = e g and exp g) >= e 1 g exp g). The projection exp g) >< exp g) is called the coherent state corresponding to g L G). In the special case g 0 we get the vacuum state exp0) >= X {0}. The linear span of the exponential vectors of M is dense in M, so that bounded operators and certain unbounded operators can be characterized by their actions on exponential vectors. The operator D : domd) M given on a dense domain domd) M containing the exponential vectors from M by Dψϕ 1, ϕ ) := ψϕ 1 + ϕ ) ψ domd), ϕ 1, ϕ M) is called compound Malliavin derivative. On exponential vectors exp g) with g L G), one gets immediately D exp g) = exp g) exp g) 1) The operator S : doms) M given on a dense domain dom S) M containing tensor products of exponential vectors by SΦϕ) := ϕ ϕφ ϕ, ϕ ϕ) Φ doms), ϕ M) is called compound Skorohod integral. One gets Dψ, Φ M = ψ, SΦ M ψ domd), Φ doms)) ) 6

7 Sexp g) exp h)) = exp g + h) g, h L G)) 3) For more details we refer to [10]. Let T be a linear operator on L G) with T 1. Then the operator ΓT) called second quantization of T is the uniquely determined) bounded operator on M fulfilling Clearly, it holds ΓT)exp g) = exp Tg) g L G)). ΓT 1 )ΓT ) = ΓT 1 T ) 4) ΓT ) = ΓT) It follows that ΓT) is an unitary operator on M if T is an unitary operator on L G). In [1] we proved. LEMMA 1.1 Let K 1, K be linear operators on L G) with property K 1K 1 + K K = 1. 5) Then there exists exactly one isometry ν K1,K from M to M = M M with Further it holds ν K1,K exp g) = expk 1 g) expk g) g L G)) 6) ν K1,K = ΓK 1 ) ΓK ))D 7) at least on domd) but one has the unique extension). The adjoint ν K 1,K of ν K1,K is characterized by and it holds ν K 1,K exp h) exp g)) = expk 1h + K g) g, h L G)) 8) ν K 1,K = SΓK 1) ΓK )) 9) 7

8 From K 1, K we get a transition expectation ξ K1 K : M M M, using ν K1,K and the lifting ξ K 1 K may be interpreted as a certain splitting cf. []). The property 5) implies K 1 g + K g = g g L G)) 10) Let U, V be unitary operators on L G). If operators K 1, K satisfy 5), then the pair ˆK 1 = UK 1, ˆK = V K fulfill 5). Here we explain fundamental scheme of beam splitting [8]. We define an isometric operator V α,β for coherent vectors such that V α,β exp g) = exp αg) exp βg) with α + β = 1. This beam splitting is a useful mathematical expression for optical communication and quantum measurements []. As one example, take α = β = 1 in the above formula and let K 1 = K be the following operator of multiplication on L G) K 1 g = 1 g = K g g L G)) We put ν := ν K1,K, then we obtain ν exp g) = exp ) 1 g exp 1 g) g L G)). 11) Another example is given by taking K 1 and K as the projections to the corresponding subspaces H 1, H of the orthogonal sum L G) = H 1 H. In [1] we used the first example in order to describe a teleportation model where Bob performs his experiments on the same ensemble of the systems like Alice. Further we used a special case of the second example in order to describe a teleportation model where Bob and Alice are spatially separated cf. section 5 of [1]). 8

9 Previous results on teleportation Let us review some results obtained in [1]. We fix an OS {g 1,..., g } L G), operators K 1, K on L G) with 5), an unitary operator T on L G), and d > 0. We assume TK 1 g k = K g k k = 1,..., ), 1) K 1 g k, K 1 g j = 0 k =j; k, j = 1..., ), 13) Using 11) and 1) we get K 1 g k = K g k = 1. 14) From 1) and 13) we get K g k, K g j = 0 k j ; k, j = 1,..., ). 15) The state of Alice asked to teleport is of the type ρ = λ s Φ s Φ s, 16) s=1 where Φ s = c sj exp ak 1 g j ) exp 0) j=1 ) c sj = 1; s = 1,..., j 17) and a = d. One easily checks that exp ak 1 g j ) exp 0) ) j=1 and exp ak g j ) exp 0) ) j=1 are OS in M. The set {Φ s; s = 1,..., } makes the -dimensional Hilbert space H 1 defining an input state teleported by Alice. We may include the vaccum state exp 0) to define H 1, however we take the -dimensional Hilbert space H 1 as above because of computational simplicity. In order to achieve that Φ s ) s=1 is still an OS in M we assume c sj c kj = 0 j k ; j, k = 1,..., ). 18) j=1 9

10 Denote c s = [c s1,..., c s ] C, then c s ) s=1 is an COS in C. Let b n ) n=1 be a sequence in C, with properties b n = [b n1,..., b n ] b nk = 1 n, k = 1,..., ), 19) b n, b j = 0 n j ; n, j = 1,..., ). 0) ow, for each m, n = 1,..., ), we have unitary operators U m, B n on M given by B n exp ak 1 g j ) exp 0) = b nj exp ak 1 g j ) exp 0) j = 1,..., ) 1) U m exp ak 1 g j ) exp 0) = exp ak 1 g j m ) exp 0) j = 1,..., ) ).1 A perfect teleportation Then Alice s measurements are performed with projection F nm = ξ nm ξ nm n, m = 1,..., ) 3) given by ξ nm = 1 b nj exp ak 1 g j ) exp 0) > exp ak 1 g j m ) exp 0), j=1 where j m := j + mmod ). 4) One easily checks that ξ nm ) n,m=1 is an OS in M. Further, the state vector ξ of the entangled state σ = ξ ξ is given by ξ = 1 exp ak 1 g k ) exp 0) exp ak g k ) exp 0). 5) k In [1] we proved the following theorem. 10

11 THEOREM.1 For each n, m = 1,...,, define a channel Λ nm by Λ nm ρ) := tr 1 F nm 1) ρ σ) F nm 1) tr 13 F nm 1)ρ σ)f nm 1) ρ normal state on M) 6) Then we have for all states ρ on M with 16) and 17) Λ nm ρ) = ΓT)U m B n )ργt)u mb n ) 7) REMARK. Using the operators B n, U m, ΓT), the projections F nm are given by unitary transformations of the entangled state σ : F nm = B n U m ΓT )) σ B n U m ΓT )) 8) or ξ nm = B n U m ΓT )) ξ. If Alice performs a measurement according to the following selfadjoint operator F = n,m=1 z nm F nm with {z nm n, m = 1,..., } R {0}, then she will obtain the value z nm with probability 1/. The sum over all this probabilities is 1, so that the teleportation model works perfectly. Before stating some fundamental results in [1] for non-perfect case, we note that our perfect teleportation is obviously treated in general finite Hilbert spaces H k k = 1,, 3) same as usual one []. Moreover, our teleportation scheme can be a bit generalized by introducing the entagled state σ 1 on H 1 H defining the projections {F nm } by the unitary operators B n, U m. We here discuss the perfect teleportation on general Hilbert spaces H k k = 1,, 3). Let { ξ k j ; j = 1,, } be COS of the Hilbert space H k k = 1,, 3). Define the entangled states σ 1 and σ 3 on H 1 H and H H 3, respectively, such as σ 1 = ξ 1 ξ 1, σ 3 = ξ 3 ξ 3 11

12 with ξ 1 1 j=1 ξ1 j ξ j and ξ 3 1 j=1 ξ j ξ3 j. By a sequence {b n = [b n1,..., b n ]; n = 1,, } in C with the properties 19) and 0), we define the unitary operator B n and U m such as B n ξ 1 j b nj ξ 1 jn, j = 1,, ) andu m ξ j ξ j mn, j = 1,, ) with j m j + m mod ). Then the set {F nm ; n, m = 1,, } of the projections of Alice is given by F nm = B n U m ) σ 1 B n U m ) and the teleportation channels {Λ nm ; n, m = 1,, } are defined as Λ nm ρ) := tr 1 F nm 1)ρ σ 3 ) F nm 1) tr 13 F nm 1) ρ σ 3 )F nm 1) ρ normal state on H 1 ). Finally the unitary keys {W nm ; n, m = 1,, } of Bob are given as W nm ξj 1 = b njξj m 3, n, m = 1,, ) by which we obtain the perfect teleportation Λ nm ρ) = W nm ρw nm. The above perfect teleportation is unique in the sense of unitary equivalence.. A non perfect teleportation We will review a non-perfect teleportation model in which the probability teleporting the state from Alice to Bob is less than 1 and it depends on the density parameter d may be the energy of the beams) of the coherent vector.there, when d = a tends to infinity, the probability tends to 1. Thus the model can be considered as asymptotically perfect. 1

13 Take the normalized vector η := with γ := γ exp ag k ) 9) k=1 )1 1 = 1 + 1)e d and we replace in 6) the entangled state σ by )e a σ := ξ ξ 30) ξ := ν K1,K η) = γ exp ak 1 g k ) exp ak g k ) k=1 Then for each n, m = 1,...,, we get the channels on any normal state ρ on M such as Λ nm ρ) := tr 1 F nm 1)ρ σ) F nm 1) tr 13 F nm 1)ρ σ) F nm 1) )1 31) Θ nm ρ) := tr 1 F nm F + )ρ σ) F nm F + ) tr 13 F nm F + )ρ σ)f nm F + ), 3) where F + = 1 exp 0) exp 0), i.e.., F + is the projection onto the space M + of configurations having no vacuum part; One easily checks that M + := {ψ M exp 0) exp 0) ψ = 0} Θ nm ρ) = F + Λ nm ρ)f + ) 33) tr F + Λnm ρ)f + that is, after receiving the state Λ nm ρ) from Alice, Bob has to omit the vacuum. From Theorem.1 it follows that for all ρ with 16) and 17) Λ nm ρ) = F +Λ nm ρ)f + tr F + Λ nm ρ)f + ). 13

14 This is not true if we replace Λ nm by Λ nm, namely, in general it does not hold Θ nm ρ) = Λ nm ρ) In [1] we proved the following theorem. THEOREM.3 For all states ρ on M with 16) and 17) and each pair n, m = 1,..., ), we have Θ nm ρ) = Γ T) U m B n )ργ T) U mb n ) or Θ nm ρ) = Λ nm ρ) 34) and p nm ρ) = tr 13 F nm F + )ρ σ)f nm F + ) = n,m n,m ) 1 e d 1 + 1)e d. 35) That is, the model works only asymptotically perfect in the sense of condition E). With other words, in the case of high density or energy) of the considered beams the model works perfectly. 3 Main results The tools of the teleportation model considered in section.1 are the entangled state σ and the family of projections F nm ) n,m=1. In order to have a more handy model, in section.. we have replaced the entangled state σ by another entangled state σ given by the splitting of a superposition of certain coherent states 30). In addition now we are going to replace the projectors F nm by projectors F nm defined as follows. F nm := B n U m ΓT) ) σ B n U m ΓT) ) 36) In order to make this definition precise we assume, in addition to ), that is holds: U m exp0) = exp0) m = 1,..., ) 14

15 Together with ) that implies U m expak 1 g j ) = expak 1 g j m ) m, j = 1,..., ) 37) Formally we have the same relation between σ and F nm like the relation between σ and F nm cf. Remark.). Further for each pair n, m = 1,..., we define channels on normal states on M such as ) ) Fnm F + ρ σ) Fnm F + Θ nm ρ) := tr 1 38) p nm ρ) where cf. 33), and 34)). ) ) p nm ρ) := tr 13 Fnm F + ρ σ) Fnm F + 39) In section 4, we will prove the following theorem. THEOREM 3.1 For each state ρ on M with 16), and 17), each pair n, m= 1,..., ) and each bounded operator A on M it holds ) tr Θnm ρ)a trλ nm ρ)a) e d 1 e d ) + ) + 40) p nmρ) 1 e d ) 41) From Theorem.1 and e d 0 d + ), the theorem 3.1 means that our modified teleportation model works asymptotically perfect case of high density or energy) in the sense of conditions E1), and E). In order to obtain a deeper understanding of the whole procedure we are going to discuss another representation of the projectors F nm and of the channels Θ nm. Starting point is again the normalized vector η given by 9). From 14) we obtain O K 1g k = g k, 4) 15

16 where O f denotes the operator of multiplication corresponding to the number or function) f Furthermore 13) implies O f ψ := fψ ψ L G)) 43) O f K 1 g k, O f K 1 g j = 0 k j) 44) From 4), and 44) follows that we have a normalized vector η given by η := Γ O K ) γ 1 η = k=1 exp a ) K 1 g k 45) REMARK 3. In the case of Example?? we have η = η ow let V be the unitary operator on M M characterized by V expf 1 ) expf )) ) 1 = exp f 1 f ) exp ) 1 f 1 + f ) f 1, f L G)) 46) On easily checks V expf 1 ) expf )) ) ) 1 1 = exp f 1 + f ) exp f f 1 ) f 1, f L G)) 47) REMARK 3.3 V describes a certain exchange procedure of particles or energy) between two systems or beams cf. [13]) 16

17 ow, using 1), 30), 45), and 47), resp. 46) one gets ξ = ν K1,K η) = 1 ΓT))V exp0) η ) 48) ξ = 1 ΓT))V η exp0) ) 49) and it follows σ = ξ ξ = 1 ΓT))V exp0) exp0) η η )1 ΓT))V ) 50) σ = 1 ΓT))V η η exp0) exp0) )1 ΓT))V ) 51) From the definition of F nm 36) and 50) it follows F nm = B n U m ) V exp0) exp0) η η )B n U m ) V ) 5) Using 51), and 5) we obtain ) ) Fnm F + ρ σ) Fnm F + where n, m = 1,..., ) = X nm 1) W nm ρ η η exp0) exp0) )W nm X nm 1) 53) X nm := B n U m ) V n, m = 1,..., ) W nm := exp0) exp0) η η F + ) V 1) B n U m ΓT))1 V ) 54) X nm and consequently X nm 1 are unitary operators. For that reason we get from 53) ) ) tr 13 Fnm F + ρ σ) Fnm F + n, m = 1,..., ) = tr 13 W nm ρ η η exp0) exp0) )W nm 55) and ) ) tr 1 Fnm F + ρ σ) Fnm F + = tr 1 W nm ρ η η exp0) exp0) )W nm 56) 17

18 ow from 38), 39), 55) and 56) it follows p nm ρ) = tr 13 W nm ρ η η exp0) exp0) )W nm 57) W nm ρ η η exp0) exp0) )Wnm Θ nm ρ) = tr 1 tr 13 W nm ρ η η exp0) exp0) )Wnm 58) According to 57,58) and 54), the procedure of the special teleportation model can be expressed in the following steps. 18

19 Step 0 initial state s in ρ) = ρ η η exp0) exp0) ρ the unknown state Alice want to teleport exp0) exp0) vacuum state, Bobs state at the beginning Step 1 Transformation according to 1 V that means: splitting of the state η η Step Transformation according to Bn U m ΓT) Step 3 Transformation according to V 1 exchange of particles or energy) between the first and the second part of the system Step 4 measurement according to exp0) exp0) η η F + checking for - first part in the vacuum? - in the third part is no vacuum? - second part reconstructed? Final state s fin ρ) = Wnms inρ))w nm tr 13 W nms in ρ))w nm ow from 57) we get Θ nm ρ) = tr 1 s fin ρ). Thus theorem 3.1 means that in the case of high density or energy) d we have approximately ρ with 16), and 17)) tr 1 s fin ρ) = ΓT)U m B n )ργt)u mb n ) The proof of theorem 3.1 shows that we have even more, namely it holds approximately) s fin ρ) = exp0) exp0) η η ΓT)U m B n )ργt)u mb n ) 59) 19

20 Adding in our scheme the following step Step 5 Transformation 1 1 ΓT)U m B n ) that means Bob uses the key provided to him) Then s fin ρ) will change into the new final state exp0) exp0) η η ρ Summarizing one can describe the effect of the procedure for large d!) as follows: At the beginning Alice has e. g., can control) a state ρ, and Bob has the vacuum state e. g., can control nothing). After the procedure Bob has the state ρ and Alice has the vacuum. Furthermore the teleportation mechanism is ready for the next teleportation e. g. η η is reproduced in the course of teleportation). We have considered three different models cf. sections.1,.,.3). Each of them is a special case of a more general concept we are going to describe in the following: Let H 1, H be dimensional subspaces of M + such that ΓT) maps H 1 onto H, and H 1 is invariant with respect to the unitary transformations B n, U m n, m = 1,..., ). Further let σ 1, σ be projections of the type σ k = ξ k ξ k, ξ k H 1 M 0 ) H M 0 ) k = 1, ) where M 0 is the orthogonal complement of M +, e. g., M 0 is the onedimensional subspace of M spanned by the vacuum vector exp0). ow for each n, m = 1,..., and each pair σ 1, σ we define a channel Ω σ 1,σ nm from the set of all normal states ρ on H 1 into the set of all normal states on M + F σ 1 nm nm ρ) := tr F +) ρ σ )F σ 1 nm F +) 1 tr 13 F σ 1 nm F + )ρ σ ) F σ 1 nm F + ) Ω σ 1σ 0

21 where F σ 1 nm := B n U m ΓT ))σ 1 B n U m ΓT )) In this paper we have considered the situation where H 1 is spanned by the OS and H is spanned by the OS expak 1 g k ) exp0) ) k=1 expak g k ) exp0) ) k=1 Further the model discussed in section. corresponds to the special case σ 1 = σ = σ, e. g. Λ nm = Ω σσ nm n, m = 1,..., ) perfect in the sense of conditions E1) and E)). The model discussed in section. corresponds to the special case σ 1 = σ σ = σ, e. g. Θ nm = Ω σ σ nm perfect in the sense of E1), and only asymptotically perfect in the sense of E)). Finally the model from this section corresponds to the special case σ 1 = σ = σ, e. g. Θ nm = Ω σ σ nm non-perfect, neither E) nor E1) hold, but asymptotically perfect in the sense of both conditions) 4 Proof of Theorem 3.1 From 14) we get exp ak s g j ) exp0) = e a 1 s = 1, ; j = 1,..., ) 60) 1

22 exp ak s g j ) = e a s = 1, ; j = 1,..., ) 61) Using 46), 60) and 61) one easily checks V exp ak 1 g j )) exp0) exp ak 1 g k ) 6) ) ) 1 [ ) ) = e a 1 e a a a exp K 1 g j g k ) exp K 1 g j + g k ) exp a ) )] a K 1 g k ) exp K 1g k k, j = 1,..., ) LEMMA 4.1 Put for j, k = 1,..., α jk := exp0) η, V exp ak 1 g j ) exp0) exp ak 1 g k ) ) Then it holds for j, k = 1,..., α jk = 1 e a )e a) 1 γ k j) 63) α jj = )1 1 e a γ 64) Proof: We have exp0), expf) = 1 f L G)) 65) Using 6), 65), and 45) we get for j, k = 1,..., α jk = ) ) 1 e a 1 e a γ s=1 ) ) ] a exp ak1 g s, exp K 1 g k ) We have [ exp ) a ak 1 g s ), exp K 1 g j + g k ) 66) expf 1 ), expf ) = e f 1,f f 1, f L G)) 67)

23 Using 13) and 67) we obtain ) 0 = exp ak1 g s ) a, exp K 1 g j + g k ) exp ak1, g s, exp a K 1g k s j) )) 68) From 61), 66), 67), and 68) it follows α jk = e a 1 ) e a e a ) 1 γ ) n e a K 1 g j, K 1 g j +g k ) e a K 1 g j, K 1 g k 69) ow 13) and 14) implies For that reason 63), and 64) follow from 69). K 1 g j, K 1 g k = 1 δ jk 70) In the following we fix a pair n, m {1,..., }. REMARK 4. Without loss of generality we can assume B n = 1 71) That we can explain as follows: Using 57,58), 59), and 54) we obtain in the case 71) Θ km ρ) = Θ nm B k ρb k) k = 1,..., ) p km ρ) = p nm B k ρb k) k = 1,..., ) On the other hand from theorem.1 follows that in the case 71 ) for all states ρ with 16) and 17) it holds Λ km ρ) = Λ nm B k ρb k) k = 1,..., ) Finally it is easy to show that B k ρb k fulfills 16), and 17) if the state ρ fulfills 16) and 17). For that reasons theorem 3.1 would be proved if we could prove 40), and 41) on the assumption that we have 71). 3

24 U m ow from 30), 49), and 37) we get γ ΓT))V η exp0) ) = exp ak 1 g k ) expak g k m ) k=1 7) LEMMA 4.3 Put for s = 1,..., β s := exp0) exp0) η η ) V 1)1 U m ΓT))1 V ) Ψ s η exp0) exp0) s = 1,..., ) 73) Then it holds β s = γ +e a ) 1 1 e a exp0) η ) c sj exp ak g k ) j=1 k=1 Proof: From 17), 7), and 73) we get β s = j=1 c sj 1 e a ) c sj exp ak g j m ) j=1 74) γ [ exp0) exp0) η η )V exp ak 1 g j ) exp0) k=1 exp ak 1 g k ) )] exp ak g k m ) 75) Further we have exp0) exp0) η η )V exp ak 1 g j )) exp0) exp ak 1 g k ) = exp0) η exp0) η, V exp ak 1 g j ) exp0) exp ak 1 g k ) ) j, k = 1,..., ) 76) Using Lemma 4.1, 75), and 76) we obtain β s = γ + γ )1 1 e a exp0) η c sj exp ak g j m ) j=1 1 e a )e a) 1 exp0) η 4 ) c sj exp ak g k m ) j k j

25 That implies 74). Since ow we put one easily checks Ψ 0 := 1 exp ak 1 g j ) exp0) 77) F + exp ak r g k ) = j=1 F + = 1 exp0) exp0), )1 1 e a exp ak r g k ) exp0) r = 1, ; k = 1,..., m) 78) Using 77), and 78) we obtain ) F + exp ak g k ) k=1 = = )1 1 e a Um ΓT) Ψ 0 79) )1 1 e a ΓT)Um Ψ 0 Using the same arguments we get ) F + c sj exp ak g j m ) j=1 = = )1 1 e a U m ΓT) Ψ s s = 1,... 80), ) )1 1 e a ΓT)U m Ψ s 81) Finally we have exp0) exp0) η η F + ) V 1) = 1 1 F + ) exp0) exp0) η η )V 1 8) 5

26 Using 54), 71), 79), 80), and Lemma 4.3 one easily checks the following equality. = γ W nm Ψ s η exp0) ) s = 1,..., ) 83) ) 1 e a exp0) η ΓT)U m Bn ) ) ) ) 1 e a Ψ s + e a c sj Ψ0 For that reason we have the following Lemma LEMMA 4.4 For each bounded operator A on M and s = 1,..., it holds ϑ s A) : = W nm Ψ s η exp0)), 1 1 A)W nm Ψ s η exp0) ) ) γ ) [ ) = 1 e a 1 e a ΓT)Um B n Ψ s, AΓT)U m B Ψ s ) ) +e a 1 e a c sj ΓT)Um Bn Ψ s, AΓT)U m Bn Ψ 0 +e a +e a j 1 e a j j ) ) c sj ΓT)Um Bn Ψ 0, AΓT)U m Bn Ψ s j c sj ΓT)U m B n Ψ 0, AΓT)U m B n Ψ 0 ow from 16) we get = ρ η η exp0) exp0) λ s Ψ s η exp0) Ψ s η exp0) 84) s=1 On the other hand Ψ s η exp0) ) s=1 is an OS because Ψ s) s=1 is an OS. for that reason from 57,58), 84), and Lemma 4.4 with A = 1 it follows ) [ γ ) ) p nm ρ) = 1 e a 1 e a + e a λ s c sj s=1 s=1 + ) e a 1 e a λ s c sj Ψ s, Ψ 0 + c sj Ψ s, Ψ 0 ) ] 85) s=1 j=1 6

27 As expak 1 g j ) exp0) ) j=1 is an OS we can calculate easily For that reason from 85) follows Ψ s, Ψ 0 = 1 k=1 c sk γ p nm ρ) = e a + e a ) [ ) 1 e a 1 e a ))] 1 e a ) + λ s s=1 c sj j=1 86) Further we have s λ s = 1 and c sj j j=1 c sj c sk j k csj k + c ) sk 87) Using 86), 87) and the definition of γ cf. 9 )) we can estimate p nmρ) 1 ) γ = 1 e a 1 1 e a 1 + 1)e a) 1 e a 1 1 e a That implies 41). ) 1 e a ) e a ) + λ s c sj e a + e a s ) λ s s j j c sj e a + e a ) )e a) + e a + ) ) ) ) e a + ) 7 )) 1 e a 1 γ 4 )) 1 e a

28 LEMMA 4.5 We use the notation ϑ s A) from Lemma 4.4. Then for each bounded operator A on M and s = 1,..., it holds Z s A) : = ϑ s A) p nm ρ) ΓT)U mbn Ψ s, AΓT)U m Bn Ψ s e a ) + ) + 1 e a Proof: Using Lemma 4.4 and the estimation ΓT)U m B n Ψ k, AΓT)U m B n Ψ r A k, r = 0,..., ) We get γ Z s A) A γ + A +e a j ) 1 e a ) 1 e a c sj ] ) pnm ρ)) 1 1 [ ) 4 pnm ρ)) 1 e a 1 e a ) c sj j Because of 86) it follows ) 1 e a Z A ) 1 e a + λ s c sj e a + e a 1 e a s j ) e a 1 e a c sj + e a c sj j j + ) 1 e a + λ s c sj e a + )) e a 1 e a s j )) 1 8

29 Using 87) we get ) 1 e a ) 1 e a + λ s c sj e a + e a s j e a ) + ) 1 e a 1 e a )) 1 and 1 e a e a e a 1 e a 1 e a ) + λ s s j ) + ) ) c sj + e a j j c sj e a + e a c sj )) 1 e a That proves Lemma 4.5. We have the representation 84) of ρ η η exp0) exp0) as a mixture of orthogonal projections. Thus from 56) and 57,58) we get with the notation ϑ s A) from Lemma 4.4 ) tr Θnm ρ)a = λ s ϑ s A) p nm ρ)) 1 s On the other hand from Theorem.1 follows tr Λ nm ρ)a) = s λ s ΓT)U m B n Ψ s, AΓT)U m B n Ψ s Consequently we have with notation Z s A) from the Lemma 4.5 ) tr Θnm ρ)a tr Λ nm ρ)a) λ s Z s A) s For that reason 40) follows from Lemma 4.5, and s That completes the proof of Theorem 3.1. λ s = 1. 9

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