aa r X i v : . [ qu a n t - ph ] J u l SEPARABLE STATES WITH UNIQUE DECOMPOSITIONS
KIL-CHAN HA AND SEUNG-HYEOK KYE
Abstract.
We search for faces of the convex set consisting of all separable states,which are affinely isomorphic to simplices, to get separable states with unique decom-positions. In the two-qutrit case, we found that six product vectors spanning a fivedimensional space give rise to a face isomorphic to the 5-dimensional simplex with sixvertices, under suitable linear independence assumption. If the partial conjugates ofsix product vectors also span a 5-dimensional space, then this face is inscribed in theface for PPT states whose boundary shares the fifteen 3-simplices on the boundary ofthe 5-simplex. The remaining boundary points consist of PPT entangled edge statesof rank four. We also show that every edge state of rank four arises in this way. If thepartial conjugates of the above six product vectors span a 6-dimensional space thenwe have a face isomorphic to 5-simplex, whose interior consists of separable stateswith unique decompositions, but with non-symmetric ranks. We also construct a faceisomorphic to the 9-simplex. As applications, we give answers to questions in theliterature [8, 9], and construct 3 ⊗ d ≥
3, we also show that ( d + 1)-dimensional subspaces give rise to facesisomorphic to the d -simplices, in most cases. Introduction
The notion of quantum entanglement is one of the main current research subjectsin quantum information theory as well as quantum physics, where the fundamentalquestion is, of course, how to distinguish entangled states from separable ones [39].One of useful criteria is to use the notion of partial transpose, as it was observed byChoi [12] and Peres [33]: The partial transpose of a separable state is still positive,that is, positive semi-definite, but the converse is not true. Even though we have amore complete criterion using positive linear maps [15, 21], it is not so easy to applythis criterion due to the current stage of poor understanding on positive linear maps.We denote by S the convex set of all separable states on M m ⊗ M n , where M m denotesthe algebra of all m × m matrices over the complex field. A state is called separable ifit is a convex combination of pure product states. Therefore, every separable state isof the form(1) ̺ = k X i =1 λ i | x i ⊗ y i ih x i ⊗ y i | , Date : November 6, 2018.1991
Mathematics Subject Classification.
Key words and phrases. separable, entanglement, positive partial transpose, face, simplex, rank,length, unique decomposition.partially supported by NRFK 2012-0002600 and NRFK 2012-0000939. ith unit product vectors | x i ⊗ y i i := | x i i ⊗ | y i i ∈ C m ⊗ C n and positive numbers λ i satisfying P ki =1 λ i = 1. A non-separable state is called entangled. The partialtranspose ̺ Γ of a state ̺ is obtained by taking the transpose in the first component.Then the partial transpose of a separable state in (1) is given by ̺ Γ = k X i =1 λ i | ¯ x i ⊗ y i ih ¯ x i ⊗ y i | , which is still positive, where | ¯ x i denotes the vector whose entries are the complexconjugates of the corresponding entries of the vector | x i . The product vector | ¯ x ⊗ y i is called the partial conjugate of | x ⊗ y i . We denote by T the convex set of all stateswhose partial transposes are positive. The positive partial transpose (PPT) criteriontells us that the relation S ⊂ T holds. These two convex sets coincide if and only if mn ≤
6, as it was shown by Woronowicz [41], Choi [12] and Horodecki’s [21].We note that the boundary between separable states and entangled ones consistsof faces of the convex set S , and so it is very important to understand the whole facialstructures of S . First of all, we note that every extreme point of S is a pure productstate | x ⊗ y ih x ⊗ y | . The next step to understand the facial structures of S is naturallyto find faces of S which are affinely isomorphic to simplices. We call those simplicialfaces . We recall that a convex subset F of a convex set C is said to be a face of C isthe following property hold: x, y ∈ C, (1 − t ) x + ty ∈ F for some t ∈ (0 ,
1) = ⇒ x, y ∈ F. We note that every point p in a convex set determines a unique face in which p isan interior point with respect to the relative topology given by the affine manifoldgenerated by the face. This is the smallest face containing the point p . By definitionof face, it is clear that the following two statements for a separable state ̺ of the form(1) are equivalent: • ̺ determines a simplicial face. • The decomposition (1) of ̺ is unique.It was shown by Kirkpatrick [26] that if {| x i i} is pairwise distinct up to constantmultiplications and {| y i i} is linearly independent then the decomposition (1) is unique.This result was also found by Alfsen and Shultz [1] independently in the context of facialstructures. See also [2] for further progresses. Recently, Chen and Djokovi´c [9] extendedthis result to show that if product vectors in (1) with k ≤ m + n − | x i ⊗ y i i are the only product vectors in the span of them. This clearly impliesthat the separable states determined by these vectors have unique decompositions. Wenote that this result covers the cases when the ranks of separable states do not exceed m + n −
2. The main purpose of this paper is to search for separable states with uniquedecompositions whose ranks exceed the number m + n − n the two-qutrit case, it is well known that generic five dimensional subspaces of C ⊗ C have exactly six product vectors {| x i ⊗ y i i} , and their orthogonal complementshave no product vectors. We show that these six product vectors give rise to a simplicialface isomorphic to the 5-simplex ∆ whose interior consists of separable states of rankfive. The interiors of the six maximal faces, isomorphic to the 4-simplex ∆ , alsoconsist of separable states of rank five. The interiors of the next fifteen maximal faces,isomorphic to the 3-simplex ∆ , consist of separable states of rank four, and so on. Weconsider the set F of all PPT states whose ranges are contained in the span of {| x i ⊗ y i i} such that the ranges of partial conjugates are contained in the span of {| ¯ x i ⊗ y i i} . Thisis a face of the convex set T consisting of all PPT states [16]. If the partial conjugatesof six product vectors also span the 5-dimensional space, then we see that the face ∆ (we use the same notion with the simplex itself) is inscribed in F so that fifteen faces∆ are located on the boundary of F . With a mild additional condition, we also showthat the interior of six maximal faces ∆ are contained in the interior of F . Remainingboundary of F consists of PPT entangled edge states of rank four, which are extremepoints of F . We also show that every edge state of rank four arises in this way. Fromthis, we see that every PPT entangled edge state of rank four is the difference of scalarmultiples of two separable states with rank five.We have another byproducts with this picture. We see that if we take two boundarypoints of F so that they are ‘outside’ of different maximal faces, then the convex sumsof these two points intersect separable states in ∆ . This gives an affirmative answer tothe Problem 1 in [9] which asks if a sum of two extreme PPT entangled states may beseparable. For a given entangled state, it is an important problem to find the nearestseparable state. See [34]. For generic 3 ⊗ ̺ .In the above discussion, it may happen that the partial conjugates of six productvectors span the 6-dimensional space. In this case, we also have a simplicial faceisomorphic to the 5-simplex ∆ . It is interesting to note that the interior points ̺ ofthis face have unique decompositions but have asymmetric ranks: rank ̺ = rank ̺ Γ .It turns out that the convex combinations of them and the edge states of rank fiveconstructed in [29] give rise to PPT states ̺ of type (5 , ̺ = 5 andrank ̺ Γ = 9. PPT states of such type were missing in a numerical search [30], but theexistence has been noticed in [20]. We give explicit examples of such states.The length of a separable state ̺ is defined by the minimum number k for whichthe expression (1) is possible. P. Horodecki [22] showed that the length of an m ⊗ n separable state does not exceed ( mn ) . The notion of length was defined in [14], whereit was shown that the length is invariant under the operation of partial transpose, andso the length may be strictly greater than the rank of the state. See also [36, 40]. Ourpicture also gives concrete examples. Any separable state ̺ in the interior of ∆ has ength six, while rank of ̺ is five. The rank of ̺ Γ may be five or six. Therefore, thereexists a state ̺ whose length is greater than the rank of the state even in the case ofrank ̺ = rank ̺ Γ . But the lengths of these examples do not exceed mn . Now, it isnatural to ask if there exists an m ⊗ n separable state of length exceeding the number mn . In a recent paper [8], Chen and Djokovi´c showed that such a separable state existswhen ( m − n − >
1, and conjectured [8, Conjecture 10] that a certain numbermay be the maximum length. By this conjecture, the length of 3 ⊗ of S is induced by the face F of T , in the sense that ∆ is the intersection of S and a face F of T . We call such faces of S induced ones, as itwas introduced in [11]. Note that the maximal faces ∆ are not induced. We constructa simplicial face isomorphic to the 9-simplex ∆ with ten extreme points. This face isturned out to be non-induced. Our construction gives an example of a 3 ⊗ ,In the 2 ⊗ n cases, the results in [1, 9, 26] give us simplicial faces isomorphic to the( n − n ≥ n +1 choices of vectors in C ⊗ C n also give riseto simplicial faces isomorphic to the n -simplex ∆ n , under suitable linear independenceassumption. In this case, we do not have the pictures as in the 3 ⊗ n .We note that the convex hull of {| x i ⊗ y i ih x i ⊗ y i | : i = 1 , , . . . , n } is a simplexif and only if they are linearly independent in the real vector space of all Hermitianmatrices. It should be noted that this convex hull need not to be a face. In thiscontext, we begin this paper to search for conditions on the family {| x i ⊗ y i i} for which {| x i ⊗ y i ih x i ⊗ y i |} is linearly independent in the next section. In Section 3, we considerthe 3 ⊗ S in Section 5, and construct a 3 ⊗ . Finally, we consider the 2 ⊗ n cases in Section 6, and discuss briefly higher dimensional cases and related problemsin the final section.2. linear independence of pure product states In this section, we investigate to what extent pure product states may be linearlyindependent. For this purpose, the notions of generalized unextendible product basis and general position are very useful. These two notions play central roles for thedescription of 3 ⊗ {| x i ⊗ y i i} is said be a generalized unextendible product basis if there is no product ector in the orthogonal complement of the span of {| x i ⊗ y i i} . It was shown in [35]that the set {| x i ⊗ y i i ∈ C m ⊗ C n : i = 1 , , . . . , k } with k ≥ m + n − { , , . . . , k } = I ∪ J , at leastone of {| x i i : i ∈ I } or {| y j i : j ∈ J } spans the whole spaces.On the other hand, a family of product vectors {| x i ⊗ y i i ∈ C m ⊗ C n : i ∈ I } issaid to be in general position if for any J ⊂ I with | J | ≤ m the set {| x j i : j ∈ J } islinearly independent, and for any K ⊂ I with | K | ≤ n the set {| y k i : k ∈ K } is linearlyindependent, where | J | denotes the cardinality of the set J . It is clear that productvectors {| x i ⊗ y i i : i = 1 , , . . . , k } with k ≥ m + n − k = m + n − C ⊗ C . Forsix product vectors in C ⊗ C , there exists a generalized unextendible product basiswhich is not in general position. See the last paragraph of Section 3 for such a concreteexample. Therefore, these two notions are different in general.It was shown in Lemma 28 of [9] that if {| x i ⊗ y i i ∈ C m ⊗ C n : i = 1 , , . . . , k } with k ≤ m + n − | x i ⊗ y i i are the only product vectors in the span of them. It is clear that if a set ofproduct vectors is linearly independent then the corresponding pure product states arealso linearly independent, but the converse is not true. Therefore, this result tells usthat they determines a simplicial face. In the 3 ⊗ , as it was explained in Introduction. Focusing on thelinear independence, we have the following: Proposition 2.1.
Let ≤ k ≤ m + n − . If k product vectors {| x i ⊗ y i i : i = 1 , , . . . , k } are in general position in C m ⊗ C n , then they are linearly independent.Proof. We may assume that k > n . Suppose that P ki =1 c i | x i i ⊗ | y i i = 0. For anyvector | ω i in the orthogonal complement of the space span {| y i i : 1 ≤ i ≤ n − } ⊂ C n ,we see that h ω | y i i = 0 (1 ≤ i ≤ n − , h ω | y j i 6 = 0 ( n ≤ j ≤ k ) , since product vectors | x i ⊗ y i i are in general position. We define a linear map T ω from C m ⊗ C n to C m by T ω ( | x i ⊗ | y i ) = h ω | y i x . Then we have T ω k X i =1 c i | x i i ⊗ | y i i ! = 0 ⇐⇒ k X j = n c j h ω | y j i| x j i = 0 . Note that 2 ≤ k − n + 1 ≤ m . Since k − n + 1 vectors | x j i are linearly independentand h ω | y j i 6 = 0 for n ≤ j ≤ k , we see that all c j = 0 for n ≤ j ≤ k . This implies that n − i =1 c i | x i i ⊗ | y i i = 0, and thus we see that c i = 0 for 1 ≤ i ≤ n −
1. Consequently,we get c i = 0 for all 1 ≤ i ≤ k . This completes the proof. (cid:3) We note that a 3 ⊗ m + n − m = n = 3.We have seen that product vectors in general position give rise to linearly independentpure product vector states whenever the number of vectors is less than m + n . Thefollowing proposition extends this. Proposition 2.2.
Let ≤ k ≤ m + 2 n − . If k product vectors S := {| z i i = | x i ⊗ y i i : i = 1 , , . . . , k } are in general position in C m ⊗ C n , then the corresponding pure product states arelinearly independent.Proof. From Proposition 2.1, any m + n − S are linearly inde-pendent. So we may assume that m + n − < k ≤ m + 2 n −
3. We consider twosubspaces X = span {| x i i : 1 ≤ i ≤ m − } ⊂ C m , Y = span {| y i i : m ≤ i ≤ n + m − } ⊂ C n . Choose two vectors | φ i ∈ C m and | ψ i ∈ C n such that | φ i ∈ X ⊥ and | ψ i ∈ Y ⊥ . Wenote that(2) n + m − ≤ i ≤ k = ⇒ h x i | φ ih y i | ψ i 6 = 0 , since product vectors of S are in general position.Now, we suppose that P ki =1 c i | z i ih z i | = 0. Then we have that0 = k X i =1 c i | x i ⊗ y i ih x i ⊗ y i | φ ⊗ ψ i = k X i = n + m − c i h x i | φ ih y i | ψ i| x i ⊗ y i i . Since any m + n − S are linearly independent, we see that c i = 0for all n + m − ≤ i ≤ k from the condition (2). On the other hand, we know that n + m − {| z i ih z i | : 1 ≤ i ≤ n + m − } are linearly independentsince product vectors {| z i i : 1 ≤ i ≤ n + m − } are linearly independent. Consequently,we see that c i = 0 for all 1 ≤ i ≤ k . This completes the proof. (cid:3) As for the product vectors which form a generalized unextendible product basis,we have the following:
Proposition 2.3.
Suppose that six product vectors {| x i ⊗ y i i : i = 1 , , . . . , } in C ⊗ C are pairwise distinct, and form a generalized unextendible product basis. Thenthe corresponding six pure product states are linearly independent. roof. Suppose P i =1 c i | z i ih z i | = 0. Consider two sets X = {| x i , | x i , | x i} and Y = {| y i , | y i , | y i} from product vectors. By assumption, at least one of them, say X , spans the whole space C . Then, we see the followingdim(span( Y )) ≥ , dim(span( Y ∪ { y i } )) = 3 , for each i = 1 , ,
3, by assumption. Therefore, for each i = 1 , ,
3, we can choose twovectors | y i i and | y i i in Y such that y i / ∈ span {| y i i , | y i i} . We denote the element of Y \ {| y i i , | y i i} by | y i i . Now, for each i = 1 , ,
3, we choose two vectors | φ i i and | ψ i i such that | φ i i ∈ (span( X \ {| x i i} )) ⊥ , | ψ i i ∈ (span {| y i i , | y i i} ) ⊥ . We note that h x i | φ i ih y i | ψ i i 6 = 0 because dim( X ) = 3 and y i / ∈ span {| y i i , | y i i} . There-fore, we have that0 = X k =1 c k | z k ih z k | φ i ⊗ ψ i i = c i h x i | φ i ih y i | ψ i i| x i ⊗ y i i + c i h x i | φ i ih y i | ψ i i| x i ⊗ y i i for each i = 1 , ,
3. Since | x i ⊗ y i i is not parallel to | x i ⊗ y i i , we can conclude that c = c = c = 0.Now, we choose a vector | ψ i ∈ C and | y k i ∈ Y = {| y i , | y i , | y i} such that h y k | ψ i = 0 and h y k | ψ ih y k | ψ i 6 = 0 , where {| y k i , | y k i , | y k i} = Y . This is possible because dim(span( Y )) ≥
2. Then wechoose a vector | φ i ∈ C such that h x k | φ ih x k | φ i 6 = 0. Consequently, we see that c k = c k = 0 from the following equation0 = X i =4 c i | z i ih z i | φ ⊗ ψ i = c k h x k | φ ih y k | ψ i| x k ⊗ y k i + c k h x k | φ ih y k | ψ i| x k ⊗ y k i . Finally, we see that c k = 0. This completes the proof. (cid:3) Very recently, it was shown in [13] that if a separable state ̺ in (1) is not uniquelydecomposed then there exists a subset I ⊂ { , , . . . , k } with the cardinality | I | ≥ {| x i i : i ∈ I } + dim span {| y i i : i ∈ I } ≤ | I | + 1 . If k ≤ m + n − k ≤ m + n − {| x i i ⊗ | y i i} in (1) is in general positionthen every subset I violates this inequality, and so we get the unique decomposition.This recovers the result in [9]. On the other hand, if k > m + n − m + n in most cases, and the inequality is not violated inthese cases. This means that we need another arguments for unique decompositionsof separable states whose ranks exceed m + n −
2. The authors are grateful to ScottCohen for useful discussions on the above result. . Two qutrit separable states with unique decompositions
In this section, we restrict our attention to the two-qutrit case to look for separablestates with unique decompositions. We will concentrate on the rank five separablestates, since states of rank up to four are covered by the results in [1, 9, 26], as it wasdiscussed already. We begin with the following simple observation, as it was discussedin Introduction.
Proposition 3.1.
Suppose that a finite family P of product vectors in C m ⊗ C n hasthe following two properties: (A) The corresponding family {| z ih z | : z ∈ P} of pure product states is linearlyindependent, (B) If a product vector | x ⊗ y i belongs to the span of P then it is parallel to a vectorin P .Then the convex hull C P of {| z ih z | : z ∈ P} is a simplicial face of S .Proof. By the condition (A), we see that C P is a simplex. To see that it is a face of S , suppose that ̺ , ̺ ∈ S and ̺ t := (1 − t ) ̺ + t̺ belongs to C P . Then, it is clearthat the range space of ̺ := P | w i ih w i | is contained in the range space of ̺ t , whichis again contained in the span of P . By the condition (B), we have | w i i ∈ P , and so itfollows that ̺ ∈ C P . Similarily, we can show that ρ ∈ C P . Therefore, C P is a face of S . (cid:3) We note that generic four dimensional subspaces of C ⊗ C have no product vectors,and their orthogonal complements have exactly six product vectors(4) | z i i = | x i ⊗ y i i , i = 1 , , . . . , , which form a generalized unextendible product basis. By Proposition 2.3 and Proposi-tion 3.1, we see that a convex combination of product states | z i ih z i | with i = 1 , , . . . , of the convex set S of all sepa-rable states. This face is affinely isomorphic to the 5-simplex with six extreme points.The state ̺ = 16 X i =1 | z i ih z i | is located at the center of the simplicial face ∆ . We note that any interior points of thesimplex ∆ give rise to examples of separable states whose lengths are strictly greaterthan the maximum of the ranks of themselves and partial transposes, if the partialconjugates of (4) also span the 5-dimensional space. Existence of such separable stateshas been claimed in [8] without explicit construction.From now on throughout this section, we suppose that the partial conjugates of (4)also span the 5-dimensional space. Note that the kernel of any PPT entangled edgestates of rank four have this property. We consider a maximal face ∆ of ∆ which is he convex hull of five product states determined by five product vectors among (4),say, {| z i , . . . , | z i} . We first consider the case when both these five product vectorsand their partial conjugates are linearly independent, and so they span the rangesof ̺ and ̺ Γ0 , respectively. Any interior point σ of ∆ is the convex combination of {| z i ih z i | : i = 1 , , . . . , } with nonzero coefficients, and so the range spaces of ̺ and σ coincide. Therefore, there is ǫ > σ − ǫ ̺ is still positive. We applythe same argument to the partial conjugates of ̺ and σ to see that there is ǫ > σ − ǫ̺ with ǫ = min { ǫ , ǫ } is a PPT state when normalized. We note thatthis argument cannot be applied to a maximal face ∆ of ∆ whose interior point hasthe range space with five linearly independent product vectors. In this case, the rangespace of an interior point of ∆ is a proper subspace of an interior point of ∆ . Wenote that σ − ǫ̺ is entangled, because the range space has only six product vectorsin (4) whose convex combination ∆ does not contain σ − ǫ̺ . We take the largestnumber ǫ > ρ = σ − ǫ̺ is of PPT. By the maximality of ǫ , we see that the rank of ρ is less than 5. Since everyPPT state whose rank is less than or equal to 3 is separable by [23], we see that ρ is ofrank four. By the results in [6, 35], we conclude that ρ is a PPT entangled edge statewhich is an extreme point in the convex set T of all PPT states.Before going further, we remind the readers of the facial structures of the convexset T . For a pair ( D, E ) of subspaces of C m ⊗ C n , it is easy to see that the set τ ( D, E ) := { ρ ∈ T : R ρ ⊂ D, R ρ Γ ⊂ E } is a face of T , which is possibly empty. It was shown in [16] that every face of T is inthis form for a pair ( D, E ) of subspaces. It was also shown that the interior is given byint τ ( D, E ) = { ρ ∈ T : R ρ = D, R ρ Γ = E } . It is unknown which pairs of subspaces give rise to a nontrivial face of T , except forthe case of m = n = 2. See [16].In the above situation, let D be the span of the product vectors in (4) and E the span of their partial conjugates. Then ̺ is an interior point of τ ( D , E ). Wenote that ∆ = τ ( D , E ) ∩ S , and the state ρ given by (5) is an extreme point of τ ( D , E ). We have shown that the interior of the maximal face ∆ of ∆ is containedin the interior of the face τ ( D , E ). If {| z i , . . . , | z i} is linearly dependent and spansa proper subspace of R ̺ , then the above process is not possible. If we take an interiorpoint σ of the face ∆ then the line segment from ̺ to σ cannot be extended withinthe convex set T . This means that the maximal face ∆ of ∆ lies on the boundary ofthe face τ ( D , E ) of T . We summarize the above discussion in general situations. heorem 3.2. Suppose that a finite family P of product vectors in C m ⊗ C n givesrise to the simplicial face C P of S . Then, for a subfamily Q of P , the following areequivalent: (i) span Q = span P and span ¯ Q = span ¯ P , where ¯ P denotes the set of partialconjugates of members in P , (ii) int C Q ⊂ int τ (span P , span ¯ P ) , (iii) A line segment from an interior point of C P to an interior point of C Q can beextended within the convex set T , to get PPT entangled states. In Proposition 3.1, we note that the condition (B) is not necessary to get simplicialfaces, by considering the face ∆ in the above discussion. In Section 5, we considerthose cases in a systematic way. Theorem 3.2 tells us that span Q $ span P if and onlyif C Q is on the boundary of τ (span P , span ¯ P ).We note that if six product vectors (4) are in general position then they satisfy thecondition (A) by Proposition 2.2. Furthermore, every five of them are also in generalposition, and so satisfy the condition (i) by Proposition 2.1. Therefore, the simplex ∆ is inscribed in the face τ ( D , E ) of T . The boundary of τ ( D , E ) consists of fifteen3-simplices from the boundary of ∆ and extreme PPT entangled edge states of rankfour. We also note that interior points of the 3-simplices are separable states of rankfour. Unfortunately, the authors could not determine if every generalized unextendibleproduct basis consisting of six product vectors in C ⊗ C is in general position or not,when they span a 5-dimensional space. Figure 1.
The triangle, edges, vertices represent the 5-simplex, 4-simplices and 3-simplices consisting of separable states, respectively. Thetriangle is inscribed in the round convex body τ ( D , E ) so that the ver-tices are on the boundary of τ ( D , E ). The boundary points of τ ( D , E )which is not on the vertices represent PPT entangled edge states of rankfour.Now, we take two points ρ , ρ on the boundary of τ ( D , E ) which are locatedoutside of different maximal faces. Then it is clear that the line segment between ρ and ρ touches the simplex ∆ . Since ρ and ρ are extreme PPT states, we see thata convex combination of two extreme PPT entangled states may be separable. Thisgives an affirmative answer to Problem 1 of [9]. Actually, we have a stronger result, aswe will see in Corollary 3.5. o far, we have seen that six product vectors in general position which span a5-dimensional space give rise to PPT entangled edge states of rank four by (5). Weproceed to show that every edge state ρ of rank four arises in this way. To do this, wefirst note that Ker ρ has exactly six product vectors which are in general position by[6, 35]. Therefore, we have the following: Proposition 3.3.
For any ⊗ PPT entangled edge state ρ of rank four, there existsa ⊗ PPT entangled edge state of rank four whose range is orthogonal to the rangeof ρ . Now, let ρ be a 3 ⊗ σ of rank four whose range is contained in Ker ρ . ThenKer σ again has exactly six product vectors {| x i ⊗ y i i : i = 1 , , . . . } in general positionwhich generate Ker σ . Note that Ker σ contains the range of ρ .We note that the following relationTr[ ρ σ ] = Tr[( ρ σ ) Γ ] = Tr[ ρ Γ σ Γ ] = Tr[ σ Γ ρ Γ ]holds for PPT states ρ and σ . We also note that if Tr ( ρσ ) = 0 for positive ρ and σ then ρσ = 0. Therefore, we may conclude that if the range of ρ is contained in Ker σ then the range of ρ Γ is also contained in Ker σ Γ , as it was already observed in [23]. Wealso note that the easy relation | x ⊗ y i ∈ Ker σ ⇐⇒ | ¯ x ⊗ y i ∈ Ker σ Γ holds for PPT states σ . See [27]. Therefore, we conclude that Ker( σ Γ ) is spanned bysix product vectors {| ¯ x i ⊗ y i i : 1 , , . . . , } , since they are also in general position. Now,the face τ (Ker σ, Ker σ Γ ) of T contains the 5-simplex ∆ which is the convex hull of sixproduct vectors | x i ⊗ y i i . We also note that interior of ∆ is contained in the interiorof the face τ (Ker σ, Ker σ Γ ), and ρ is on the boundary of the face τ (Ker σ, Ker σ Γ ).Therefore, we have shown that every PPT entangled edge state arises from six productvectors in general position spanning 5-dimensional space, and is of the form (5). Theorem 3.4.
Every two-qutrit PPT entangled edge state of rank four is the differenceof scalar multiples of two separable states with rank five.
We also have the following corollary in the relation with the question in [9] men-tioned before.
Corollary 3.5.
For any ⊗ PPT entangled edge state ρ of rank four, there existsa PPT entangled edge state ρ of rank four such that parts of the line segment between ρ and ρ are separable. Now, suppose that ρ is a PPT state of rank five. In generic cases, the range of ρ has exactly six product vectors {| x i ⊗ y i i} which determines a simplicial face ∆ . It is lear that ρ is separable if and only if the linear equation X i =1 λ i | x i ⊗ y i ih x i ⊗ y i | = ρ, X i =1 λ i = 1has a solution ( λ , . . . , λ ) with 0 ≤ λ i ≤
1. If this is not the case, we consider thelinear equation X i =1 λ i | x i ⊗ y i ih x i ⊗ y i | = (1 − µ ) ̺ + µρ, X i =1 λ i = 1 , ≤ λ i , µ ≤ λ i and µ . If ̺ is expressed as the linear combinationof pure product states, then this equation has infinitely many solutions. But, it hasa unique solution ( λ , . . . , λ , µ ) under the condition that exactly one λ i is zero. Thissolution represents the maximal face ∆ which is the nearest from ρ among six maximalfaces of ∆ . This suggests a method to find the nearest separable state to ρ , but wecould not determine if this is really the nearest face among all faces of S .We illustrate the above discussion with the PPT state ρ defined by ρ = 13(1 + b + 1 /b ) · · · · · · · b · · · · · ·· · /b · · · · ·· · /b · · · · · · · · · · · · · · · · b · ·· · · · · b · ·· · · · · · /b · · · · · · · , where b is a positive real number with b = 1, and · denotes 0, as it was constructed in[19]. Note that ρ is a PPT entangled edge state with rank ρ = rank ρ Γ = 4. If b = 2,then this is just the first example of 3 ⊗ ρ has only following six product vectors:(6) | x ⊗ y i =(1 , √ b, t ⊗ (1 , − √ b , t , | x ⊗ y i = (1 , −√ b, t ⊗ (1 , √ b , t , | x ⊗ y i =(0 , , √ b ) t ⊗ (0 , , − √ b ) t , | x ⊗ y i = (0 , , −√ b ) t ⊗ (0 , , √ b ) t , | x ⊗ y i =( √ b, , t ⊗ ( − √ b , , t , | x ⊗ y i = ( −√ b, , t ⊗ ( 1 √ b , , t . We denote by | z i i the normalization of product vector | x i ⊗ y i i . Then it is also easy to seethat P = {| z i i : i = 1 , , . . . , } is in general position, and so it satisfies the conditions(A) and (B) of Proposition 3.1. Furthermore, any five choice of them satisfies thecondition (i) of Theorem 3.2. For each k = 1 , , . . . ,
6, we denote by F k the convex hullof {| z i ih z i | : | z i i ∈ P , i = k } . Then these F k ’s are maximal faces which are isomorphic o ∆ with the center ρ k = P i =1 i = k | z i ih z i | of rank five. We can show thatmax { ǫ : ρ k − ǫ̺ ∈ T } = 15 , k = 1 , , . . . , , where ̺ = P i =1 | z i ih z i | . Therefore, we see that σ k = 54 ( ρ k − ̺ ) , k = 1 , , . . . , ̺ and ρ k of rank five. Note that the range space of σ k is orthogonal to the range spaceof ρ . We also have σ ( t ) := tσ i + (1 − t ) σ j = 5 − t | z i ih z i | + 6 t − | z j ih z j | + 524 X k =1 k = i, j | z k ih z k | . By considering the coefficients of | z i ih z i | and | z j ih z j | , we see that the state σ ( t ) on theline segment between σ i and σ j is separable if ≤ t ≤ . The converse is also ture,because σ ( ) and σ ( ) are interior points of maximal faces F j and F i of S , respectively.We denote by {| i i : i = 1 , , } the usual orthonormal basis of C . If we replacethe vector | x i in (6) by | x i = | i then it is straightforward to see that they forma generalized unextendible product vectors. It is cleat that they are not in generalposition. It is also easy to see that they span the 6-dimensional space. It would beinteresting to know whether a generalized unextendible product vectors consisting ofsix vectors in C ⊗ C are in general position when they span the 5-dimensional space.4. Simplicial faces of asymmetric type and optimality of decomposableentanglement witnesses
So far, we have considered the five dimensional subspaces which have exactly sixproduct vectors (4) whose partial conjugates also span the five dimensional spaces.In this section, we consider the case when the partial conjugates of (4) span the six-dimensional space. As a byproduct, we construct PPT states of type (9 , ̺ is of type ( p, q ) if rank ̺ = p and rank ̺ Γ = q .We begin with the 3 ⊗ ̺ θ of type (5 ,
5) constructed in[29]. For a fixed positive real number b with b = 1, define ̺ θ = e iθ + e − iθ · · · − e iθ · · · − e − iθ · b · − e − iθ · · · · ·· · b · · · − e iθ · ·· − e iθ · b · · · · ·− e − iθ · · · e iθ + e − iθ · · · − e iθ · · · · · b · − e − iθ ·· · − e − iθ · · · b · ·· · · · · − e iθ · b ·− e iθ · · · − e − iθ · · · e iθ + e − iθ , here − π < θ < π and θ = 0. First of all, we note that the kernel of ̺ θ is spanned bythe following four vectors(7) | w i =(1 , , , , , , t , | w i =(0 , b, e iθ , , , , t , | w i =(0 , , , , b ; 0 , e iθ , t , | w i =(0 , , e iθ ; 0 , , b, , t . From this, it is easy to see that the 5-dimensional range space of ̺ θ has exactly sixproduct vectors, and the kernel has no product vectors. We list up all of them withthe temporary notation ω = √ b e i θ :(8) | z i = (1 , ω, t ⊗ ( ω, − , t = ( ω, − , ω , − ω, , , t , | z i = ( − , ω, t ⊗ ( ω, , t = ( − ω, − , ω , ω, , , t , | z i = (0 , , ω ) t ⊗ (0 , ω, − t = (0 , , , ω, − , ω , − ω ) t , | z i = (0 , − , ω ) t ⊗ (0 , ω, t = (0 , , , − ω, − , ω , ω ) t , | z i = ( ω, , t ⊗ ( − , , ω ) t = ( − ω, , ω ; 0 , , − , , ω ) t , | z i = ( ω, , − t ⊗ (1 , , ω ) t = ( ω, , ω ; 0 , , − , , − ω ) t . Define the separable state ̺ sep by ̺ sep = 12 b X i =1 | z i ih z i | = · · · − · · · − · b · − e − iθ · · · · ·· · b · · · − e iθ · ·· − e iθ · b · · · · ·− · · · · · · − · · · · · b · − e − iθ ·· · − e − iθ · · · b · ·· · · · · − e iθ · b ·− · · · − · · · . It is clear that ̺ sep is of type (5 ,
6) and has a unique decomposition by Proposition 2.3and Proposition 3.1. In this way, we have separable states with unique decompositionswhich are of asymmetric type.Now, we consider the PPT state defined by ̺ = 12 ( ̺ sep + ̺ θ ) . t is clear that ̺ is of rank five. It is also easy to see that the partial transpose ̺ Γ θ · · · − e iθ · · · − e − iθ · b · − e − iθ · · · · ·· · b · · · − e iθ · ·· − e iθ · b · · · · ·− e − iθ · · · θ · · · − e iθ · · · · · b · − e − iθ ·· · − e − iθ · · · b · ·· · · · · − e iθ · b ·− e iθ · · · − e − iθ · · · θ is of rank nine. In this way, we have parameterized examples of 3 ⊗ ̺ Γ of type (9 , D of all decomposable positivelinear maps and PPT states, to consider the dual face of the PPT states ̺ Γ given by( ̺ Γ ) ′ = { φ ∈ D : h ̺ Γ , φ i = 0 } . Recall that the bilinear pairing is defined by h ̺, φ i = Tr ( ̺C t φ ), where C φ is the Choimatrix of the linear map φ . Because ̺ Γ is of type (9 , ̺ Γ ) ′ consists of completely copositive maps. Therefore, we conclude that every mapin the dual face ( ̺ Γ ) ′ gives rise to a decomposable entanglement witness. Recall thatthe kernel of ( ̺ Γ ) Γ = ̺ is spanned by four vectors in (7). Therefore, entanglementwitnesses we constructed are the partial transposes of positive matrices supported onsubspaces of Ker ̺ . We consider the entanglement witness W θ given by W θ = | w ih w | + 1 b X i =2 | w i ih w i | ! Γ = · · · e iθ · · · e − iθ · b · · · · · ·· · b · · · · ·· · b · · · · · e − iθ · · · · · · e iθ · · · · · b · ·· · · · · b · ·· · · · · · b · e iθ · · · e − iθ · · · , which is the partial transpose of a positive matrix supported on Ker ̺ .Recall that an entanglement witness W is called optimal [31] if it detects a maximalset of entanglement with respect to the set inclusion. A decomposable entanglementwitness W is the partial transpose of a positive matrix with the support E . Necessaryconditions for optimality of a decomposable witness W were found in [28] in terms of E : If W is optimal then E satisfies the following:(i) E has no product vector.(ii) E ⊥ has a product vector.(iii) The convex hull of {| z ih z | Γ : z ∈ E } is a face of D under the correspondence φ C φ . n the 2 ⊗ n cases, it was shown in [4] that (i) is equivalent to the optimality of W .With extra necessary conditions (ii) and (iii) above, it was shown in [28] that (i) is notsufficient for optimality in general. By the discussion above, it is now clear that thewitness W θ satisfies the above three conditions: Ker ̺ has no product vector; (Ker ̺ ) ⊥ is spanned by the six product vectors in (8); The convex hull of {| z ih z | Γ : z ∈ Ker ̺ } coincides with ( ̺ Γ ) ′ , and so, it is actually an exposed face of D . In the earlier versionof this paper, the authors wrote that the witness W θ is optimal. But they found a faultin the reasoning after submission, as it was also observed in [3] where it was shownthat W θ is not optimal. Therefore, we may conclude that the above three conditionsare not sufficient in general for optimality of decomposable entanglement witnesses.5. Induced and non-induced faces
In Section 3, we have seen two kinds of simplicial faces of the convex set S . Theface ∆ arising from six product vectors in (6) is the intersection of S with a face ofthe larger convex set T consisting of all PPT states. On the other hands, its maximalface ∆ has no such face of T , because the smallest face of T containing ∆ has theintersection ∆ with S . We distinguish these two cases.A face F of the convex set S is said to be induced by a face τ ( D, E ) of T , or τ ( D, E ) induces the face F of S if the relation F = τ ( D, E ) ∩ T , int F ⊂ int τ ( D, E )holds for a pair (
D, E ) of subspaces in C m ⊗ C n , as it was introduced in [11]. Thispair is uniquely determined by the above relation. If a face F of S is induced by a faceof T , then we just say that it is an induced face. A face τ ( D, E ) induces a face of S if and only if it has a separable state in its interior. It was shown in [11] that this isthe case if and only if the pair ( D, E ) satisfies the range criterion [22], that is, thereexist product vectors | x i ⊗ y i i such that D = span {| x i ⊗ y i i} and E = span {| ¯ x i ⊗ y i i} .This tells us how the partial converse of the range criterion works. In order to testfor a separable state ̺ to determine an induced face, we denote by P ( ̺ ) the set of allproduct vectors | z i such that the state | z ih z | belongs to the face determined by ̺ , and P ( ̺ ) the set of all product vectors | x ⊗ y i ∈ R ( ̺ ) such that | ¯ x ⊗ y i ∈ R ( ̺ Γ ) . It wasshown in [18] that P ( ̺ ) ⊂ P ( ̺ ) holds in general, and the equality holds if and only if ̺ determines an induced face of S . Proposition 5.1.
Let P = {| x i ⊗ y i i : 1 , , . . . , n } be a family of normalized productvectors. Then the convex hull C P of {| z ih z | : z ∈ P} is an induced simplicial faceof S if and only if P satisfies the condition (A) in Proposition together with thefollowing: (C) If a product vector z = | x ⊗ y i belongs to the span of P and | ¯ x ⊗ y i belongs tothe span of ¯ P then it is parallel to a vector in P . roof. Suppose that C P is an induced simplicial face. Then we have the condition(A) together with the relation(9) C P = S ∩ τ (span P , span ¯ P ) . To prove the condition (C), suppose that a product vector | z i satisfies the assumptionof (C). Then | z ih z | ∈ τ (span P , span ¯ P ) ⊂ C P by (9). Since C P is a simplex, we seethat the condition (C) holds.For the converse, suppose that the conditions (A) and (C) holds. If | z i = | x ⊗ y i isa product vector and ̺ = | z ih z | belongs to τ (span P , span ¯ P ) then we have | x ⊗ y i ∈ R ̺ ⊂ span P , | ¯ x ⊗ y i ∈ R ̺ Γ ⊂ span ¯ P . By the condition (C), we see that ̺ ∈ C P . This shows one direction of the inclusionof (9). Since the other direction is true always, we have the relation (9) which tells usthat C P is an induced face. Finally, C P is a simplex by the condition (A). (cid:3) In the remaining of this section, we construct a simplicial face which is not induced.We modify the example (6) as follows:(10) (1 , √ b, t ⊗ (1 , √ b , t , (1 , −√ b, t ⊗ (1 , − √ b , t , (0 , , √ b ) t ⊗ (0 , , √ b ) t , (0 , , −√ b ) t ⊗ (0 , , − √ b ) t , ( √ b, , t ⊗ ( 1 √ b , , t , ( −√ b, , t ⊗ ( − √ b , , t , where we still retain the condition b = 1. We see that these six product vectors spanthe 6-dimensional space whose orthogonal complement is spanned by(11) √ b | i ⊗ | i − √ b | i ⊗ | i , √ b | i ⊗ | i − √ b | i ⊗ | i , √ b | i ⊗ | i − √ b | i ⊗ | i . From this, it is easy to see that there are infinitely many product vectors in the spanof product vectors (10). Since all entries are real, these vectors do not satisfy thecondition (C). Nevertheless, we show that these vectors make a simplicial face. So wewill get a simplicial face which is not induced. Actually we show that this is a subfaceof a simplicial face isomorphic to the 9-simplex.To do this, we consider the product vectors:(12) (1 , , t ⊗ (1 , , t , (1 , , − t ⊗ (1 , , − t , (1 , − , t ⊗ (1 , − , t , ( − , , t ⊗ ( − , , t . hese four product vectors span the 4-dimensional space whose orthogonal complementis spanned by | i ⊗ | i − | i ⊗ | i , | i ⊗ | i − | i ⊗ | i , | i ⊗ | i − | i ⊗ | i , | i ⊗ | i − | i ⊗ | i , | i ⊗ | i − | i ⊗ | i . From this, it is immediate to see that these product vectors satisfy the condition (A) and(B), and so determine a simplicial face. We denote by P the family of normalizationsof the above product vectors (12) together with six product vectors in (10). We notethat both P and ¯ P span the full space C ⊗ C , and so it is clear that the condition(C) does not hold. In addition to this fact, we show that C P is a simplicial face. So C P is also a simplicial face which is not induced. This will prove that every interior pointof C P gives rise to a separable state of rank nine whose length is ten. This disprove aconjecture in [8], as it was discussed in Introduction.We first show that the convex set C P is a simplex. To do this, we label thenormalized vectors in (10) by | z i , | z i , . . . , | z i , and normalized vectors in (12) by | z i , . . . , | z i . Suppose that P i =1 α i | z i ih z i | = 0. If | w i is any one of (11) then we have h z i | w i 6 = 0 for i = 7 , , ,
10. Since {| z i , . . . , | z i} is linearly independent, we see fromthe relation 0 = X i =1 α i | z i ih z i | w i = X i =7 α i h z i | w i| z i i that α i = 0 for i = 7 , , ,
10. Since {| z i , | z i , . . . , | z i} is also linearly independent,we conclude that α i = 0 for i = 1 , , . . . ,
6. Therefore, we see that {| z i ih z i | : i =1 , , . . . , } is linearly independent, and so C P is a simplex.In order to show that C P is a face, we use the duality between positive linear mapsand separable states to see that C P is a dual face of a positive linear map. We recallthe generalized Choi map Φ[ α, β, γ ] between M defined byΦ[ α, β, γ ]( X ) = αx + βx + γx − x − x − x γx + αx + βx − x − x − x βx + γx + αx for X ∈ M and nonnegative real numbers α, β and γ , as it was introduced in [10]. Werestrict our attention to the maps Φ[ α, β, γ ] satisfying the following condition0 < α < , α + β + γ = 2 , βγ = (1 − α ) , which implies that Φ[ α, β, γ ] is a positive linear map. This condition is parameterizedby α ( s ) = (1 − s ) − s + s , β ( s ) = s − s + s , γ ( s ) = 11 − s + s , with 0 < s < ∞ , s = 1. Therefore, we obtain parameterized positive mapsΦ( s ) = Φ [ α ( s ) , β ( s ) , γ ( s )] , < s < ∞ , s = 1 s in [17] and [18] (See page 333 in [17] and page 6 in [18]). We also recall that everypositive linear map Φ( s ) gives rise to the dual faceΦ( s ) ′ = { ρ ∈ S : Tr( C tΦ( s ) ρ ) = 0 } of S , where C tΦ( s ) is the transpose of the Choi matrix of the map Φ( s ) given by C Φ( s ) = X i,j =1 | i ih j | ⊗ Φ s ( | i ih j | )= α ( s ) · · · − · · · − · γ ( s ) · · · · · · ·· · β ( s ) · · · · · ·· · · β ( s ) · · · · ·− · · · α ( s ) · · · − · · · · · γ ( s ) · · ·· · · · · · γ ( s ) · ·· · · · · · · β ( s ) ·− · · · − · · · α ( s ) ∈ M ⊗ M . It is known from Theorem 3 in [18] that every product vector whose product statebelongs to the dual face Φ(1 /b ) ′ is one of the following forms(13) (¯ a , ¯ a , ¯ a ) t ⊗ ( a , a , a ) t with | a | = | a | = | a | , (0 , ¯ a , b ¯ a ) t ⊗ (0 , a , a ) t with | a | = b | a | , ( b ¯ a , , ¯ a ) t ⊗ , ( a , , a ) t with | a | = b | a | , (¯ a , b ¯ a , t ⊗ ( a , a , t with | a | = b | a | , and so the dual face Φ(1 /b ) ′ is the convex hull of pure product states correspondingto these product vectors. We note that P is contained in the dual face Φ(1 /b ) ′ . Fur-thermore, we see that P coincides with the set of product vectors in (13) with realcomponents up to scalar multiplications. Now, we consider the dual face (Φ( s ) ◦ t) ′ where Φ s ◦ t is the composition of Φ s and the transpose map t. We note that C Φ( s ) ◦ t = X i,j =1 | i ih j | ⊗ (Φ( s ) ◦ t)( | i ih j | ) = X i,j =1 | i ih j | ⊗ Φ( s )( | j ih i | ) = C ΓΦ( s ) . Thus we see that a product vector | x ⊗ y i belongs to the dual face (Φ( s ) ◦ t) ′ if andonly if | ¯ x ⊗ y i belongs to the dual face Φ( s ) ′ . From this observation, we easily see that | x ⊗ y i ∈ P if and only if | x ⊗ y i ∈ Φ(1 /b ) ′ ∩ (Φ(1 /b ) ◦ t) ′ , which coincides with thedual face of the map Φ(1 /b ) + Φ(1 /b ) ◦ t. This shows that the convex set C P is just thedual face of the positive map Φ(1 /b ) + Φ(1 /b ) ◦ t. Consequently, C P is a non-inducedsimplicial face of S with the center ̺ = 110 X i =1 | z i ih z i | , which is isomorphic the 9-simplex ∆ . e note that the method to get PPT entanglement with Theorem 3.2 works inmore general situations. Suppose that F is a face of the convex set S , which is notnecessarily isomorphic to a simplex. Take an interior point ̺ and a boundary pointof σ of F . If the range spaces of ̺ and σ coincide, then we can take ǫ > ρ = σ − ǫ̺ is positive. We note that this state ρ must be entangled. To see this, assumethat ρ is separable. Then we see that σ is the convex sum of two separable states ̺ and ρ , and so we conclude that ρ ∈ F by the definition of face. This contradictionshows that ρ must be entangled. If the range spaces of ̺ Γ0 and σ Γ also coincide thenwe can take ǫ > ρ = σ − ǫ̺ and ρ Γ = σ Γ − ǫ̺ Γ0 are positive. Thenwe see that ρ must be PPT entangled state by the same argument. Proposition 5.2.
Let F be a face of the convex set S with an interior point ̺ and aboundary point σ . If ρ = σ − ǫ̺ is positive for ǫ > then ρ is an entangled state. Ifboth ρ and ρ Γ are positive then ρ is a PPT entangled state. Now, we return to the above simplicial face C P = ∆ . For each k = 1 , , . . . , F k the convex hull of {| z i ih z i | : | z i i ∈ P , i = k } . Then these F k ’s aremaximal faces of C P , which are isomorphic to ∆ with the center σ k = 19 X i =1 i = k | z i ih z i | of rank nine. We also see that for any interior point σ of F k , the range spaces of ̺ and σ coincide, and so we can apply the above argument. Using a symbolic computationprogram, we can show that λ k := max { ǫ : σ k − ǫ̺ ∈ T } = b ) b + b ) , k = 1 , , , , , , b b + b ) , k = 7 , , , . In this way, we have PPT entangled states ρ k = σ k − λ k ρ with rank ρ k = rank ρ Γ k = 8,which are located at the boundary of T .6. qubit-qudit cases In this section, we turn our attention to the 2 ⊗ n cases, and look for simplicialfaces determined by separable states with rank n + 1, because the results in [1, 9, 26]covers separable states with rank up to n . We take n + 1 product vectors in generalposition. Then these vectors are linearly independent by Proposition 2.1, and so thecorresponding pure product states satisfy the condition (A) in Proposition 3.1. Wewill show that these vectors also satisfy the condition (C) in Proposition 5.1 under amild assumption. Consequently, the corresponding pure product states determine aninduced simplicial face as follows. heorem 6.1. Suppose that P = {| e i ⊗ f i i : i = 1 , , . . . , n +1 } is a family of normalizedproduct vectors, which are in general position in C ⊗ C n . We write | e i i = a i | e i + a i | e i ∈ C , i = 3 , , . . . , n + 1 , then we have the following: (i) If n = 2 then the face determined by P has infinitely many extreme points. (ii) Let n ≥ . In this case, if a n +1 , a k ¯ a n +1 , ¯ a k / ∈ R for each k = 3 , , . . . , n , then P determines an induced simplicial face isomorphic to the n -simplex.Proof. Write | f n +1 i = P ni =1 b i | f i i . Then none of a ij ′ s and b k ′ s is zero by assumptionof general position. Suppose that a product vector | x ⊗ y i satisfies the assumption ofcondition (C). We also write | x i = x | e i + x | e i and | y i = P ni =1 y i | f i i , with x i , y j ∈ C .Then there exist complex numbers s i , t i such that(14) | x ⊗ y i = n +1 X i =1 s i | e i ⊗ f i i = s | e ⊗ f i + s | e ⊗ f i + n X i =3 2 X j =1 s i a ij | e j ⊗ f i i + X j =1 n X k =1 s n +1 a n +1 ,j b k | e j ⊗ f k i , | ¯ x ⊗ y i = n +1 X i =1 t i | ¯ e i ⊗ f i i = t | ¯ e ⊗ f i + t | ¯ e ⊗ f i + n X i =3 2 X j =1 t i ¯ a ij | ¯ e j ⊗ f i i + X j =1 n X k =1 t n +1 ¯ a n +1 ,j b k | ¯ e j ⊗ f k i . Since {| e i ⊗ f j i : i = 1 , , j = 1 , , . . . , n } form a basis of C ⊗ C n , we can solve x i , y j and s k , t k by comparing the coefficients of basis vectors in (14),If s n +1 = 0, then we see that x y = 0 and x y = 0. It is easy to that | x ⊗ y i = s | e ⊗ f i if x = y = 0 ,s | e ⊗ f i if x = y = 0 ,s k | e k ⊗ f k i ( k = 3 , , . . . , n ) if y = y = 0 , because product vectors | e i ⊗ f i i are in general position. This shows that there existexactly n product vectors in the span of P \ {| e n +1 ⊗ f n +1 i} whose partial conjugatesbelong to the span of partial conjugates of product vectors in P .Now, we consider the case s n +1 = 0. We look at the coefficients of | e ⊗ f i and | e ⊗ f i to get x y = s n +1 a n +1 , b , x y = s n +1 a n +1 , b , ¯ x y = t n +1 ¯ a n +1 , b , ¯ x y = t n +1 ¯ a n +1 , b . We see that x i , y j are nonzero, and so we may assume that x = a n +1 , . Then we have y = s n +1 b and t n +1 = s n +1 from the first and third equalities. From the second andforth equalities, we also have x = r a n +1 , and y = s n +1 b /r for a nonzero real number . Therefore, we have(15) x = a n +1 , , x = ra n +1 , , y = s n +1 b r , y = s n +1 b , t n +1 = s n +1 for a nonzero real number r . If n = 2 then this shows that there are infinitely manyproduct vectors in the span of P whose partial conjugates belong to the span of partialconjugates of product vectors in P , and this shows that the face determined by P hasinfinitely many extreme points. Consequently, we proved the assertion (i).Next, suppose that n ≥
3. For k ≥ | e ⊗ f k i and | e ⊗ f k i to solve y k , s k , t k and s n +1 . After substituting x , x by the values in (15)with s n +1 = t n +1 , we have the following equation(16) − a n +1 , a k a n +1 , b k − ra n +1 , a k a n +1 , b k − ¯ a n +1 , a k ¯ a n +1 , b k − r ¯ a n +1 , a k ¯ a n +1 , b k y k s k t k s n +1 = , where the determinant of the 4 × r − b k ( a n +1 , a k ¯ a n +1 , ¯ a k − ¯ a n +1 , ¯ a k a n +1 , a k ) . This determinant must be zero, because the homogeneous equation (16) has a nonzerosolution by the current assumption s n +1 = 0. By the assumption of the statement (ii),we have r = 1. In this case, the matrix in (16) is of rank three, with the solution s k = t k = 0 and y k = s n +1 b k . Therefore, we have | x ⊗ y i = s n +1 | e n +1 ⊗ f n +1 i . By combining the results for the cases of both s n +1 = 0 and s n +1 = 0, P satisfiesthe condition (C) in Proposition 5.1. We already know that P also satisfies the con-dition (A) in Proposition 3.1. Therefore, P determines an induced simplicial face byProposition 5.1. (cid:3) It should be noted that every ( n + 1)-dimensional subspace of C ⊗ C n has infinitelymany product vectors, as it was shown in Lemma 10 of [27]. If every vector in P hasreal entries, then P and ¯ P coincide and so there are infinitely many product vectors inspan P = span ¯ P . The curious condition a n +1 , a k ¯ a n +1 , ¯ a k / ∈ R may reflect this fact.In the low dimensional cases, we can now classify all simplicial faces. In the caseof 2 ⊗ n with n = 2 ,
3, two notions of separability and PPT coincide, and so faces ofthe convex set S are classified in terms of pairs ( D, E ) of subspaces of C ⊗ C n . Wefirst find a necessary condition for a separable state to determine an induced simplicialface. We note that every face is induced in the 2 ⊗ ⊗ Proposition 6.2.
Suppose that a ⊗ n separable state ̺ of the form (1) determinesan induced simplicial face of the convex set S . Then {| x i i} is pairwise distinct up toscalar multiplications. In the cases of n = 2 , , P = {| x i ⊗ y i i : i = 1 , , . . . , k } is ingeneral position. roof. Assume that two of {| x i i} , say | x i and | x i , are parallel to each other. Foreach | y i ∈ span {| y i , | y i} , we consider the product state ̺ y = | x ⊗ y ih x ⊗ y | . Thenwe see that | x ⊗ y i ∈ R ̺ and | ¯ x ⊗ y i ∈ R ̺ Γ . Therefore, we see that ̺ y belongs tothe face S ∩ τ ( R ̺, R ̺ Γ ) of S determined by ̺ . This face is not a simplex because thereare infinitely many extreme points ̺ y for arbitrary | y i ∈ span {| y i , | y i} . Therefore,we see that {| x i i} is pairwise distinct up to scalar multiplications.For the second claim, it suffices to consider the the case of n = 3. If two of {| y i i} are parallel to each other then ̺ cannot determine a simplicial face by the same reasonas above. If three of them, say | y i , | y i , | y i , span a 2-dimensional space V of C , then P = {| x i ⊗ y i i : i = 1 , , } is contained in C ⊗ V , and so the face determined by P contains already infinitely many extreme points by Theorem 6.1 (i). This shows that P must be in general position. (cid:3) In the case of 2 ⊗
2, all pairs of subspaces giving nontrivial faces are classified in[16]. The possible pairs (dim D, dim E ) of natural numbers arising from non-trivialfaces are listed by (1 , , (2 , , (3 , , (3 , , (4 , , (4 , . If we take an interior point ̺ of the face τ ( D, E ) then these numbers are nothing but(rank ̺, rank ̺ Γ ). It is clear that the (1 ,
1) case gives rise to the 0-simplices consistingof single points. For the (2 ,
2) case, we take an interior point ̺ = P i =1 | z i ih z i | withproduct vectors | z i , | z i . Then we see that this face is simplicial if and only if {| z i i} isin general position. If this is the case, then we have 1-simplices. We show that these areall possible cases. To do this, suppose that a state ̺ with expression (1) is an interiorpoint of a simplicial face τ ( D, E ). The case (3 ,
3) is split into two subcases: either both D ⊥ and E ⊥ are spanned by product vectors x ⊗ y and ¯ x ⊗ y , respectively, or both ofthem are spanned by non-product vectors. In the first case, it is clear that there areinfinitely many extreme points. In the second case, we note that three product vectorsare in general position if and only if they form a generalized unextendible productbasis. Therefore, we see that there are also infinitely many extreme points in this caseby Theorem 6.1 (i). In the (3 ,
4) case, D ⊥ is spanned by a product vector, and we seethat there are infinitely many product vectors in D by Lemma 10 of [27]. All of themgive rise to extreme points of the face of τ ( D, E ).We turn our attention to the 2 ⊗ , , (2 , , (3 , , (3 , , (4 , , (4 , , (4 , , (5 , , (5 , , (6 , . by [11]. Here, we list up the cases ( p, q ) with p ≤ q by the symmetry. Suppose that ̺ of the form (1) determines a simplicial face isomorphic to the k -simplex. If k ≤
3, thenthis face must be of type ( k, k ) by the above list. We next consider the case k = 4.In this case, {| x i ⊗ y i i : i = 1 , , , } is in general position by Proposition 6.2, and so heir partial conjugates are also in general position. Therefore, both {| x i ⊗ y i i} and {| ¯ x i ⊗ y i i} are linearly independent by Proposition 2.1. This show that the face mustbe of type (4 , ,
4) is never a simplicial face. In most (4 ,
4) cases,they actually determines a simplicial faces isomorphic to the 3-simplex by Theorem 6.1(ii). For a face τ ( D, E ) of type (4 , , t ⊗ (0 , , t and (1 , z ) t ⊗ (1 , z, z ) t for z = 0 , , ω, ω , where ω is the third root of unity. It issimple to check that these five product vectors satisfy the conditions in Proposition5.1. Therefore, we have a simplicial face isomorphic to ∆ . For the (4 ,
6) and (5 , D + dim E > n for subspaces D and E of C ⊗ C n then there are infinitely many product vectors in D whose partial conjugates belong to E . Therefore, it is a simple consequence of thisresult that there is no simplicial face of type (5 , Conclusion
We analyzed the convex geometry of the convex set S consisting of all 3 ⊗ T consisting of PPT states. With this geometricreasoning, we could show that every PPT entangled edge state of rank four arises asthe difference of two separable states of rank five. For the higher dimensional cases,we recall that the maximal dimension of subspaces in C m ⊗ C n without product vectoris given by p = ( m − × ( n − , and generic ( p + 1)-dimensional subspaces contain exactly (cid:0) m + n − n − (cid:1) numbers of productvectors up to scalar multiplications. See [5, 24, 32, 37, 38]. In the 3 ⊗ p + 1 = 5 < (cid:18) m + n − n − (cid:19) to get the picture shown in Figure 1. In the 3 ⊗ p + 1 = 7 and (cid:0) m + n − n − (cid:1) = 10. If these ten product vectors are in general position then they giverise to a simplicial face isomorphic to the 9-simplex by Proposition 2.2. Six of themmust be linearly independent by Proposition 2.1, but we could not determine if sevenof them are linearly independent in general or not. If n = 4, then the the number (cid:0) m + n − n − (cid:1) = (cid:0) m +23 (cid:1) exceeds eventually the whole dimension (4 m ) . Therefore, we cannotexpect linear independence of corresponding product states as in Proposition 2.2. Itwould be interesting to know to what extent our approach may work. We also haveconstructed separable states with unique decompositions, but with asymmetric ranks.As an application, we constructed an explicit example of a 3 ⊗ , n the 2 ⊗ n cases, the situations go in a very different way, because p + 1 = n = (cid:18) m + n − n − (cid:19) for these cases. Even though we found simplicial faces with n + 1 vertices, these donot give the pictures as in Figure 1, because the corresponding product vectors arealready linearly independent. This may explain why it is very difficult to constructPPT entangled edge states of various types in the 2 ⊗ n cases. In fact, it is still openif there exists a 2 ⊗ , References [1] E. Alfsen and F. Shultz,
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