Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes

Commun. Theor. Phys. (Beijing, China) 4 (004) pp. 351 360 c International Academic Publishers Vol. 4, No. 3, September 15, 004 Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes QIAN Shang-Wu 1 and GU Zhi-Yu 1 Physics Department, Peking University, Beijing 100871, China Physics Department, Capital Normal University, Beijing 100037, China (Received December 5, 003) Abstract By using the block division method in matrix calculus, this article successfully calculate the expectation values of the generating operators and the correlation tensors for quantum network of any nodes. Thence, by means of the criterion of entanglement in terms of the covariance correlation tensor in quantum network theory, this article discusses the complete and partial separability conditions of the mixed quantum state for a quantum network of any nodes and judge the separability of a quantum state in the general case of any nodes for two examples. PACS numbers: 03.67.Mn, 03.65.Ta, 0.10.Yn Key words: covariance correlation tensor in quantum network theory, criterion of entanglement, mixed state 1 Introduction Quantum entanglement plays an essential role in quantum teleportation, quantum computation, and the modern understanding of quantum phenomena. Recently we have found that there is an intimate relationship between quantum entanglement and the knot theory in topology. In our four previous papers [1 4] we have discussed the relationship between quantum entanglement and the oriented links in knot theory. [5 7] Furthermore we have used the theory of quantum networks [8] to discuss the criterion of entanglement in terms of the covariance correlation tensor (CCT) [9] and use this criterion to solve the problem of separability of two nodes, [10] three nodes, [11,1] and four nodes. [13] We have successfully found the complete separability conditions and the general solution of a pure quantum state with real coefficients for a quantum network of any nodes. [14] In this article we shall further discuss complete and partial separability conditions of the mixed state for a quantum network of any nodes. In the case of m nodes, the product states µ = µ(1) µ() µ(m) = µ(1)µ() µ(m), µ(ν) represents one of the two possible states (ν) and (ν). There are m product states. The general expression of a pure state is [8] Ψ = m µ=1 C µ µ, m µ=1 For this pure state the density operator is ˆρ(1,,..., m) = Ψ Ψ = C µ = 1. (1) m i, ρ ij ˆPij, () ρ ij = C i C j, ˆPij = i j, Tr { ˆP mn ˆPjl } = δ ml δ nj. (3) In most interesting and practical cases C µ is a real number, i.e. C µ = C µ, µ = 1,,..., 16. (4) The general expression of the density operator of a mixed state of m nodes is k ˆρ(ν 1, ν,..., ν m ) = Ψ Ψ = Λ j Φ ν1,ν,...,νm j Φ ν1,ν,...,νm j, (5) Λ j > 0, k Λ j = 1. In Eq. (5), Φ ν1,ν,...,νm j are orthogonal normalized eigenstates of the mixed state Ψ, the number of such states is k, ν i is the ordinal number of the node i. In general Φ ν1,ν,...,νm j is not a separable state. In the case of separable mixed state, the wave function is represented by χ, and r ˆρ(ν 1, ν,..., ν m ) = χ χ = p i ϕ ν1,ν,...,ν h i ψ h+ i... ψi m ϕ ν1,ν,...,ν h i ψ h+ i... ψi m, = i=1 r p i ˆρ i (1,,..., h) ˆρ i (h + 1) ˆρ i (h + ) ˆρ i (m) (6) i=1 r i=1 p i = 1. In Eq. (6), ϕ ν1,ν,...,ν h i p i > 0, of h nodes, ψ h+1 i, ψ h+ i, and ψi m the number of states ϕ ν1,ν,...,ν h i ψ h+ i... ψi m is the orthogonal normalized eigenstates of the quantum network are normalized eigenstaes of the node h+1, node h+, and node m, respectively, is r. The number of separable nodes is g = m h, thus when h = 0

35 QIAN Shang-Wu and GU Zhi-Yu Vol. 4 the mixed state is completely separable. Reference [15] tells us that in the case of separable mixed state there exists a left unitary matrix Γ r k (Γ Γ = I k k ) whose elements are Γ ij = p i /Λ j yj i, such that p1 ϕ ν1,ν,...,ν h 1 ψ1 h+1 ψ1 h+... ψ m 1 Λ1 Φ ν1,ν,...,νm 1 p ϕ ν1,ν,...,ν h ψ h+1 ψ h+... ψ m = Γ Λ Φ ν1,ν,...,νm r k. (7) pr ϕ ν1,ν,...,ν h r ψr h+1 ψr h+... ψr m Λk Φ ν1,ν,...,νm k Using the expression of the elements Γ ij = p i /Λ j yj i of the matrix Γ r k, and expanding the right-hand side of Eq. (7) we get k ϕ ν1,ν,...,ν h i ψ h+ i... ψi m = yj Φ i ν1,ν,...,νm j, i = 1,,..., r, (8) as From Eqs. (7) and (8) we get k yj i = 1, i = 1,,..., r. (9) Ψ = ( p 1 ϕ ν1,ν,...,ν h 1 ψ1 h+1 ψ1 h+ ψ1 m, p ϕ ν1,ν,...,ν h ψ h+1 ψ h+ ψ m,..., p r ϕ ν1,ν,...,ν h r ψr h+1 ψr h+ ψr m k = ( p 1 yj 1, p yj,..., p r yj r ) Φ ν1,ν,...,νm j. (10) When we denote the separable state ϕ ν1,ν,...,ν h i ψ h+ i... ψi m by Ψi, and expand k yi j Φν1,ν,...,νm j in terms of direct product state bases into the form of pure state, equation (8) can be written as k m Ψ i = ϕ ν1,ν,...,ν h i ψ h+ i ψi m = yj Φ i ν1,ν,...,νm j = Cµ µ i, (11) m µ=1 The corresponding components of coherent vectors are C i µ = 1, i = 1,,..., r. (1) λ i (ν) = Ψ i ˆλ i (ν) ˆ1(1) ˆ1() ˆ1(ν 1) ˆ1(ν + 1) ˆ1(m) Ψ i, (13) which is the expectation value of the generating operator ˆλ j (ν) of SU(n) for the node ν, the unit operator and generating operators are ˆ1(ν) = (ν) (ν) + (ν) (ν), ˆλ1 (ν) = (ν) (ν) + (ν) (ν), ˆλ (ν) = i{ (ν) (ν) (ν) (ν) }, ˆλ3 (ν) = (ν) (ν) + (ν) (ν), (14) ν = 1,,..., m. From Eq. (11) we can solve Cµ i in terms of yj i, then we can easily obtain λ j(ν), the second rank, third rank,..., and the m-th rank CCT in terms of yj i. By using the conditions of complete or partial separability we can solve for yj i, then from the existence of a left unitary matrix Γ r k (Γ Γ = I k k ), whose elements Γ ij = p i /Λ j yj i, we can easily obtain the value of p i, and thus the expression of density operator of the separable mixed state. Conditions of Complete Separability and Partial Separability In Ref. [9] we have discussed the quantum network of m nodes for pure states, and obtained the following results ˆρ = ˆρ(1) ˆρ() ˆρ(m) + + s ν1,s ν,...,s νh lν 1,lν,,lν h =1 ν h >ν h 1 >>ν 1 s 1,s,...,s m l 1,l,...,l m=1 ν m>ν >>ν h+1 h= ˆρ(ν h+1 ) ˆρ(ν h+ ) ˆρ(ν m ) µ=1 1 h M l ν1 l ν...l νh (ν 1, ν,..., ν h ) [ˆλ lν1 (ν 1 ) ˆλ ν (ν ) ˆλ νh (ν h )] 1 m {M l 1l l m (1,,..., m) + mλ l1 (1)λ l () λ lm (m)} [ˆλ j1 (1) ˆλ j () ˆλ νh (m)] (15)

No. 3 Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes 353 with the similar definition of the correlation tensor and CCT for m bodies as that for three bodies. [6] From Eq. (15) we find the following conditions of complete separability of quantum network of m nodes for pure states: thence M l1l l m (1,,..., m) + mλ l1 (1)λ l (),..., λ lm (m) = 0, l j = 1,, 3, M lν1 l ν l νh (ν 1, ν,..., ν h ) = 0, ν h > ν h 1 > > ν 1, h =, 3,..., (m 1), (16) ˆρ = ˆρ(1) ˆρ() ˆρ(m). (17) From Eq. (15) we get the conditions of partial separability of g = m h quantum nodes for the pure states: M l1l l m (1,,..., m) + mλ l1 (1)λ l () λ lm (m) = 0, except M lµ1 l µ l µm g (µ 1, µ,..., µ m g ) 0, all other M lν1 l ν l νh (ν 1, ν,..., ν h ) = 0, ν h > ν h 1 > > ν 1, h =, 3,..., (m 1). (18) For the separable mixed state Ψ i, we only need to add a superscript i for the physical quantities appearing in Eqs. (15), (16) and (18), hence the conditions of complete separability of mixed state of a quantum network of m nodes are Thence we get M i l 1l l m (1,,..., m) + mλ l1 (1)λ l () λ lm (m) = 0, M i l ν1 l ν l νh (ν 1, ν,..., ν h ) = 0, ν h > ν h 1 > > ν 1, h =, 3,..., (m 1). (19) ˆρ = i p i ˆρ i (1) ˆρ i () ˆρ i (m). (0) The conditions of partial separability of g = m h quantum nodes for a mixed state are M i l 1l l m (1,,..., m) + mλ l1 (1)λ l () λ lm (m) = 0, except M i l µ1 l µ l µm g (µ 1, µ,..., µ m g ) 0, all other M i l ν1 l ν l νh (ν 1, ν,..., ν h ) = 0, ν h > ν h 1 > > ν 1, h =, 3,..., (m 1) (1) in this case we have ˆρ = i p i {ˆρ i (1) ˆρ i () ˆρ i (m) + ˆρ i (ν h+1 ) ˆρ i (ν h+ ) ˆρ i (ν m ) s ν1,s ν,...,s νh lν 1,lν,...,lν h =1 ν h >ν h 1 >>ν 1 1 h M l i ν1 l ν l νh (ν 1, ν,..., ν h ) [ˆλ i l ν1 (ν 1 ) ˆλ i l ν (ν ) ˆλ } i l νh (ν h )], () h = m g and ν m > ν > > ν h+1, i.e. in this case there is entanglement among m g nodes. 3 Node Interchange Relations When we interchange the nodes ν and ˆν, i.e. ν ˆν, the ordinal number µ will change to ˆµ, i.e. µ ˆµ. In this section we shall briefly review the general relation between ˆµ and µ when there is an interchange between the nodes ν and ˆν in our previous article. [14] In particular, we give the node interchange relations for the node interchange 1 k. The relation between ordinal number µ of the product state C µ and the number of nodes m is µ = µ(1) + µ() m + + µ(ν) m ν + + µ(m 1) 1 + µ(m) 0 + 1. (3) For example 00 0 = 1, 00 01 =, 00 010 = 3. Assume i < j, and let ν = i, ˆν = j, then from Eq. (3) we know µ = µ(1) + + µ(i) m i + + µ(j) m j + + µ(m) 0 + 1, (4) ˆµ = µ(1) + + µ(j) m i + + µ(i) m j + + µ(m) 0 + 1. (5) From Eqs. (4) and (5) we obtain ˆµ µ = [µ(j) µ(i)]( m i m j ). (6) Equation (6) is the required general relation between ν ˆν and µ ˆµ. From Eq. (6) we have discussed two particular cases and some related topics. When µ(j) = µ(i), then ˆµ µ, for this case there are two possible cases, one of which is µ(j) = µ(i) = 0, in this case ˆµ µ µ 0, the other is µ(j) = µ(i) = 1, in this case ˆµ µ µ 1. Since C 1 and C m are unchanged for any interchange of nodes, hence C 1 and C m are not suitable for being chosen as independent variable for any nodes. From Eqs. (4) and (5) we have µ 1 = µ 0 + m i + m j. (7) In order to refrain from choosing C 1 and C m as independent parameters, in Ref. [14] we chose the m independent parameters C µ (l) such that µ (l) = l 1 + 1, l = 1,,..., m. (8) When µ(j) µ(i), then ˆµ µ. Without loss of generality

354 QIAN Shang-Wu and GU Zhi-Yu Vol. 4 we take µ(i) = 0 and µ(j) = 1, then we have µ = µ 0 + m j = µ 1 m i, ˆµ = µ 0 + m i = µ 1 m j. (9) Equation (9) is the required node interchange relations. For the commonly used case 1 k, i.e. i = 1, j = k, equation (9) becomes µ = µ 0 + m k = µ 1, ˆµ = µ 0 + = µ 1 m k. (30) In general, we have found that all the four classes {µ 0 } {µ 1 }{µ}{ˆµ} can be classified into f = k groups, each containing g = m k members, i.e. every class has the same number of members k m k = m. These four classes {µ 0 }{µ 1 }{µ}{ˆµ} can be represented as {µ 0 } : (µ 0 ) f,g = (f ) m k + g, {µ} : (µ) f,g = (f 1) m k + g, {µ 1 } : (µ 1 ) f,g = [(f + k ) 1] m k + g, λ 1 (1) = Tr (C 1, C,..., C m) ( ) 1 0 Using Eq. (31), we can write Eq. (34) as λ 1 (1) = From Eq. (30), when node 1 node k, we have λ 1 (k) = Using Eq. (31), we can write Eq. (36) as λ 1 (k) = ( ) 1 0 {ˆµ} : (ˆµ) f,g = [(f + k ) ] m k + g, (31) f = 1,,..., k and g = 1,,..., m k. Eq. (31) we have {µ 0 } {µ} = (1,,..., ), From {ˆµ} {µ 1 } = ( + 1, +,..., m ). (3) 4 Calculation of the Expectation Value λ j of Generating Operator ˆλ j In order to use the conditions of complete separability and partial separability for the mixed state of a quantum network of m nodes, we must know the expectation value λ j of the generating operator ˆλ j for m nodes. [9 14] For m nodes we have λ 1 (1) = Tr [ˆρˆλ 1 (1) ˆ1() ˆ1(m)] = Ψ ˆρˆσ 1 ˆ1() ˆ1(m) Ψ. (33) Using Eq. (1) and expressing Eq. (33) in matrix form we have ( ) 1 0 ( 1 0 ) C 1 C C m = h=1 C h C +h. (34) µ 0,µ 1 (C µ 0C +µ 0 + C µ 1 C µ 1). (35) µ 0,µ 1 (C µ 0C m k +µ 0 + C µ 1 m kc µ 1). (36) k 1 f=1 m k g=1 Following the similar procedures of Eqs. (33) (35), we can get From node interchange relations (30) we obtain λ 3 (k) = k f=1 m k g=1 λ 3 (1) = Cj + C (f ) m k +gc (f 1) m k +g. (37) k [C (f ) m k +g + C [(f+ k ) ] m k +g ] + m h= +1 f=1 m k g=1 C h. (38) [C (f 1) m k +g + C [(f+ k ) 1] m k +g ]. (39) 5 Calculation of Correlation Tensor of Quantum Network of m Nodes Furthermore, in order to use the conditions of complete separability and partial separability for the mixed state of a quantum network of m nodes, we must also know how to calculate the correlation tensor. Here we give an example to show the procedure of the calculation of the 11 1 component K 111 of the l-th rank correlation tensor. From the definition of correlation tensor we have K 111 = Tr [ˆρˆσ 1 (1)ˆσ 1 () ˆσ 1 (l)ˆ1(l + 1)ˆ1(l + ) ˆ1(m)] = Tr [ˆρˆσ 1 lˆ1 (m l) ] = Tr (C 1, C,..., C m) ( ) l 1 0 ( 1 0 ) (m l) C 1 C C m. (40)

No. 3 Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes 355 Using the formula in matrix calculus, [16] Tr (AB) = devec A vec B. (41) Let A be an m n matrix and a j be its j-th column and a i be its i-th row, then veca is the following mn 1 vector, as devec A is the following 1 mn vector, a 1 a vec A =, devec A = (a1, a,..., a m ). (4) a n In order to calculate K 111 in Eq. (40) we let A = (C 1, C,..., C m) ( ) l, B = 1 0 ( 1 0 ) (m l) C 1 C C m. (43) In Eq. (43), (C 1, C,..., C m) is 1 m vector, as ( ) l 1 0 is l l matrix. In order to get their matrix product we must divide C 1, C,..., C m into l groups so that we can use block division method to calculate the matrix product. For this purpose we let (C 1, C,..., C m) = (S 1, S,..., S j,..., S l), (44) From Eqs. (43) (45) we readily obtain Similarly we get S j = (C (j 1) m l +1, C (j 1) m l +,..., C (j 1) m l +g,..., C (j 1) m l +m l). (45) devec A = (S l, S l 1,..., S j,..., S, S 1 ). (46) vec B = S 1 S j S l S j is the transpose of S j. From Eqs. (41), (46), and (47) we obtain K 111 = (S l, S l 1,..., S j,..., S, S 1 ), (47) S 1 S j S l l = S l +1 js j. (48) Using the similar procedures we can get all correlation tensors of the quantum network of m nodes. Actually equation (34) is also obtained by the method of block division of matrices. When l = h is an even number, using the same method as above we get K = (S l, S 1,..., S l j,..., S, S 1 ) 0 0 1 0 0 0 K 3333 = (S 1, S,..., S j,..., S l 1, S l) 0 0 1 0 0 0 S 1 S j S l S 1 S j S l, (49). (50)

356 QIAN Shang-Wu and GU Zhi-Yu Vol. 4 The correlation tensors K 111 (ν 1, ν,..., ν l ) are given by K 111 (ν 1, ν,..., ν l ) = ( 1) h (T, T 1,..., T j,..., T, T 1 ) 0 0 1 0 0 0 T 1 T j T, (51) the number of subscripts of K 111 is l, as the number of subscripts 1 of K 111 is m l 1, T j = (C j 1, C j ), T j is the transpose of T j, and ν 1 < ν < < ν l. Proof of Eq. (51) is given by Appendix. 6 Example of Complete Separability Conditions for Mixed State of a Quantum Network of m Nodes Here we give a concrete example of mixed state to illustrate the conditions of complete separability. Example Find the state of separability of the following mixed state Φ ± 1 are GHZ states of m nodes ˆρ(ν 1, ν,..., ν m ) = Λ Φ + 1 Φ+ 1 + (1 Λ) Φ 1 Φ 1, (5) Φ ± 1 = 1 ( (1) () (m) ± (1) () (m) ) = 1 [ m ± 1 ]. (53) Solution Introduce Ψ i = y i 1 Φ + 1 + yi Φ 1. (54) Substituting Eq. (53) into Eq. (54) we have Ψ i = C i 1 1 + C i m m, (55) C1 i = yi 1 y i, C i = yi 1 + y i. (56) m From our previous results about complete separably conditions, [14] we readily know that if C i m k +j = Ci j Ci m k +1 C1 i, k = 1,,..., m 1, j = 1,,..., m k, (57) then Ψ i is completely separable. From Eq. (56) we know that there are only two nonzero terms C1 i and C i for m C i m k +j, while for all other possible j and k, Ci m k +j 0, hence equation (57) is automatically satisfied, and equation (0) holds. Now let us find the nonzero terms of λ, K, M. From Sec. we know the only nonzero terms of λ are λ i 3(k), C i 1 ( ) (k 1) ( ) ( ) (m k) 1 0 1 0 λ i 3(k) = Tr (C1, i C, i..., C i m) C i. (58) C i m Now take ( ) (k 1) 1 0 A = (C1, i C, i..., C i m) Using Eqs. (41) and (59), equation (58) becomes λ i 3(k) = k ( ) 1 0, B = ( 1 0 ) (m k) C1 i C i C i m. (59) ( 1) j SjS i j i = (C1) i + (C i m) = y1y i i. (60) The possible nonzero correlation tensors are K i 3333 (whose rank l = h is an even number) and the m-th rank K i 111,

No. 3 Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes 357 which are K3333 i = Tr (C1, i 0..., 0, C i m) 0 0 1 0 0 0 C i 1 0 0 C i m = (y 1) i + (y) i (61) and K i 111 = (y i 1) (y i ). (6) The possible nonzero CCT are M3333 i (whose rank l = h is an even number) and the m-th rank M111, i which are l M3333 i = K3333 i λ i 3(k) and k=1 = (y i 1) + (y i ) (y i 1y i ) l (63) M i 111 = K i 111 = (y i 1) (y i ). (64) From the complete separability conditions we know and M i 33 = (y i 1) + (y i ) (y i 1y i ) = 0 (65) M i 111 = (y i 1) (y i ) = 0. (66) Solving Eqs. (65) and (66) we get y i 1 = 1, y i = ± 1. (67) Substituting Eq. (67) into Eq. (63) we obtain M i 3333 = (y i 1) +(y i ) (y i 1y i ) l = 0, hence all the possible nonzero terms of CCT equal zero for the values of y i 1 and y i given by Eq. (67), therefore equation (67) satisfies the complete separability conditions. From Eq. (5) we know Now we take Λ 1 = Λ, Λ = 1 Λ. (68) p 1 = p, p = 1 p. (69) Substituting Eqs. (68) and (69) into Γ ij = p i /Λ j yj i we get p p Λ (1 Λ) (Γ ij ) = 1 p 1 p. (70) Λ (1 Λ) Substituting Eq. (70) into Γ Γ = I k k we obtain 1 p 1 Λ ( ) Λ(1 Λ) 1 0 p 1 1 =. (71) Λ(1 Λ) (1 Λ) Solving Eq. (71) we get Λ = p = 1/, i.e. Since λ 1 (k) = λ (k) = 0, hence p 1 = p = 1. (7) ˆρ i (k) = 1 ˆ1(k) + 1 λi 3(k)ˆλ 3 (k) = 1 (k) (k) + (k) (k) + 1 (yi 1y i )[ (k) (k) + (k) (k) ] = 1 + yi 1y i (k) (k) + 1 yi 1y i (k) (k). (73) From Eqs. (67) and (73) we have Substituting Eqs. (7) and (74) into Eq. (0) we get ˆρ 1 (k) = (k) (k), ˆρ (k) = (k) (k). (74) ˆρ = 1 (1) () (m) (1) () (m) + 1 (1) () (m) (1) () (m). (75) 7 Example of Partial Separability Conditions for Mixed State of a Quantum Network of m Nodes Here we shall give a concrete example of mixed state to illustrate the conditions of partial separability. Example Find the state of separability of the following mixed state ˆρ(ν 1, ν,..., ν m ) = Λ Φ + 1 Φ+ 1 + (1 Λ) Φ+ Φ+, (76) Φ + 1 and Φ+ are GHZ states of m nodes, Φ + 1 = 1 ( (1) () (m) + (1) () (m) ) = 1 [ m + 1 ], Φ + = 1 ( (1) () (m 1) (m) + (1) () (m 1) (m) ) = 1 [ m 1 + ]. (77) Solution Introduce Ψ i = y i 1 Φ + 1 + yi Φ +. (78)

358 QIAN Shang-Wu and GU Zhi-Yu Vol. 4 Substituting Eq. (77) into Eq. (78) we have Ψ i = yi 1 ( m + 1 ) + yi ( m 1 + ), (79) hence C1 i = C i = yi 1, C m i = C i m 1 = yi. (80) For Eq. (80), the possible nonzero λ are λ 1 (k), which are λ 1 (k) = Tr (S 1, S,, S j,..., S l 1, S k) ( ) (k 1) 1 0 ( 1 0 ) S 1 S j S k = (S 1 S + S 3 S 4 + + S k 1S k), (81) S j = (C (j 1) m k +1, C (j 1) m k +,..., C (j 1) m k +g,..., C (j 1) m k +m k). (8) From Eqs. (80) (8) we know that the only nonzero λ 1 (k) is the term λ 1 (m), λ i 1(m) = y i 1y i. (83) Using the block division method we have λ i (k) = λ i 3(k) = 0. (84) For the possible nonzero correlation tensors, we can also use the block division method to get the desired results, e.g. when l is an even number, from Eqs. (49) and (50) we know the l-th rank correlation tensor K i = (y i 1) (y i ), (85) K i 3333 = (y i 1) + (y i ). (86) From Eq. (48) we know the possible nonzero l-th rank correlation tensors K i 111 occur when l = m 1 and l = m, which are K i 111 = y i 1y i, when l = m 1 ; K i 111 = (y i 1) + (y i ), when l = m. (87) Substituting Eq. (80) into Eq. (51) we have K 111 = ( 1) h y i 1y i. (88) From Eqs. (87) and (83) we get M i 111(1,,..., m) = K i 111(1,,..., m) K i 111(1,,..., m 1)λ i 1(m) = (y i 1) + (y i ) 4(y i 1) (y i ). (89) Using the block division method we also have M i (1,,..., m) = K i (1,,..., m) = (y i 1) (y i ). (90) Substituting Eqs. (84) and (88) into the definition formula of CCT, we have M 111 = ( 1) h y i 1y i. (91) Let M111(1, i,..., m) = M(1, i,..., m) = 0, we obtain y1 i = 1, y i = ± 1. (9) Using Eq. (9) we find some nonzero CCTs which do not include node m, such as M i 3333(ν 1, ν,... ν l ) = 1, M 1 111(1,,..., m 1) = 1, M 111(1,,..., m 1) = 1, M 1 111(ν 1, ν,..., ν l ) = ( 1) h, M 111(ν 1, ν,..., ν l ) = ( 1) h+1, ν 1 < ν <... < ν l, l = h m. (93) All the CCTs which include node m equal zero. Hence the mixed state (76) is partially separable, and the separated node is node m. From Eq. (76) we know Now we take Λ 1 = Λ, Λ = 1 Λ. (94) p 1 = p, p = 1 p. (95) Substituting Eqs. (94) and (95) into Γ ij = p i /Λ j y i j and solving Γ Γ = I k k we obtain Λ = p = 1/, i.e. p 1 = p = 1. (96) Substituting Eq. (9) into Eq. (83) we get λ (1) 1 = 1, λ () 1 = 1. (97) Using Eqs. (83), (84), and (97) we obtain ˆρ (1) (m) = 1 ˆ1(m)+ 1 λ(1) 1 (m)ˆλ 1 (m) = α + (m) α + (m), ˆρ () (m) = 1 ˆ1(m)+ 1 λ() 1 (m)ˆλ 1 (m) = α (m) α (m), (98)

No. 3 Complete and Partial Separability Conditions of Mixed State for a Quantum Network of Any Nodes 359 and ˆρ (i) (k) = ˆ1(k) 1 = 1 [ (k) (k) + (k) (k), k m. (99) From Eq. (15), the density operator of the other m 1 nodes (1,,..., m 1) for the partially separable case is ˆρ i (1,, m 1) = ˆρ i (1) ˆρ i () ˆρ i (m 1) + s ν1,s ν,...,s νh lν 1,lν,...,lν h =1 ν h >ν h 1 >>ν 1 m ν m>ν >>ν h+1 h= ˆρ i (ν h+1 ) ˆρ i (ν h+ ) ˆρ i (ν m ) 1 h M i l ν1 l ν,...,l νh (ν 1, ν,..., ν h ) [ˆλ i l ν1 (ν 1 ) ˆλ i ν (ν ) ˆλ i ν h (ν h )]. (100) Retaining nonzero terms and substituting Eq. (99) into Eq. (100) we obtain ˆρ i (1,,..., m 1) = 1 l } M3333(ν i 1, ν,..., ν l ) ˆ1(ν j ) ˆλ i 3(ν i ) + 1 { { l=1 M111(ν i 1, ν,..., ν l ) l=1 j=l+1 ˆλ i 1(ν j ) j=l+1 l i=1 i=1 } ˆλ i (ν i ) + 1 { ν=1 } ˆ1(ν) + ˆλ i 1(ν). (101) Substituting Eq. (99) and all the nonzero CCTs (93) into Eq. (101), after rather tedious calculations, finally we get Since ˆρ (1) (1,,..., m 1) = Φ + 1() Φ+ 1(), ˆρ() (1,,..., m 1) = Φ 1() Φ 1(), (10) hence from Eqs. (98) and (10) we have Φ ± 1() == 1 ( (1) () (m 1) ± (1) () (m 1) ). (103) ˆρ(1,,..., m) = p i ˆρ i (1,,..., m 1) ˆρ i (m), (104) i=1 ˆρ(1,,..., m) = 1 Φ+ 1() Φ+ 1() α+ (m) α + (m) + 1 Φ 1() Φ 1() α (m) α (m). (105) 8 Conclusion In this article we have successfully calculated the expectation values of the generating operators and the correlation tensors for quantum network of any nodes by using the block division method in matrix calculus. Therefore, by means of the criterion of entanglement in terms of the covariance correlation tensor in quantum network theory, we can easily discuss the complete and partial separability conditions of the mixed quantum state in this case and judge the separability of a quantum state of any nodes. Appendix: Proof of Eq. (51) From the matrix expression of correlation tensor we have ( ) 1 0 K 111 (ν 1, ν,, ν l ) = Tr (C 1, C,..., C m)ˆσ (m l 1) 1 ˆσ l C, (A1) C m l = h. Take Noting ˆσ ( ) 1 0 A = (C 1, C,..., C m)ˆσ (m l 1) 1 ˆσ l, B = C 1 C C m ν=1. C 1 0 0 0 0 0 0 1 0 = ( 1) = 0 0 ( 1)ˆσ 1 ˆσ 3 = ( 1) 0 0 0 0 1 0, 0 0 (A) (A3)

360 QIAN Shang-Wu and GU Zhi-Yu Vol. 4 we have From Eq. (A4) we have 0 0 ( ) (m h 1) A = ( 1) h 0 0 (C 1, C,..., C m) 1 0 0 0 = ( 1) h (C 1, C,..., C m) ( ) () 1 0 0 0 1 0 0 0 deveca = ( 1) h (T 1, T,..., T j,..., T 1, T ) 0 0 1 0 0 0. (A4) ( ) () 1 0 0 0 1 0 0 0 = ( 1) h (T, T 1,..., T j,..., T, T 1 ) 0 0 1 0 0 0 Using the method of block division, similarly we have vecb = T 1 T j T. (A5). (A6) Substituting Eqs. (A5) and (A6) into Eq. (41) we readily obtain the required result Eq. (51). References [1] S.W. Qian and Z.Y. Gu, J. Phys. A: Math. Gen. 35 (00) 3733. [] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 38 (00) 41. [3] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 41 (004) 01. [4] Z.Y. Gu, S.W. Qian, and J.S. Wang, Commun. Theor. Phys. (Beijing, China) 41 (004) 531. [5] L.H. Kauffman, Knots and Physics, nd ed., World Scientific, Singapore (1993); H. Kleinert, Path Integrals in Quantum Mechanics and Polymer Physics, World Scientific, Singapore (1990). [6] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 35 (001) 431. [7] Braid Group, Knot Theory and Statistical Mechanics, eds. C.N. Yang and M.L. Ge, World Scientific, New Jersey (1989). [8] G. Mahler and V.A. Weberrufz, Quantum Networks, nd revised and enlarged ed., Springer, Berlin, New York (1998). [9] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 39 (003) 15. [10] Z.Y. Gu and S.W. Qian, Commun. Theor. Phys. (Beijing, China) 39 (003) 41. [11] Z.Y. Gu and S.W. Qian, Commun. Theor. Phys. (Beijing, China) 40 (003) 33. [1] Z.Y. Gu and S.W. Qian, Commun. Theor. Phys. (Beijing, China) 40 (003) 151. [13] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 40 (003) 395. [14] S.W. Qian and Z.Y. Gu, Commun. Theor. Phys. (Beijing, China) 4 (004) 69. [15] S.J. Wu, X.M. Chen, and Y.D. Zhang, Phys. Lett. A75 (000) 44. [16] D.A. Turkington, Matrix Calculus and Zero-One Matrices, Cambridge University Press, Cambridge (00).