aa r X i v : . [ qu a n t - ph ] F e b The Minimum Size of Qubit UnextendibleProduct Bases
Nathaniel Johnston
Institute for Quantum Computing, University of WaterlooWaterloo, Ontario N2L 3G1, Canada [email protected]
Abstract
We investigate the problem of constructing unextendible product bases in the qubit case – thatis, when each local dimension equals 2. The cardinality of the smallest unextendible productbasis is known in all qubit cases except when the number of parties is a multiple of 4 greaterthan 4 itself. We construct small unextendible product bases in all of the remaining open cases,and we use graph theory techniques to produce a computer-assisted proof that our constructionsare indeed the smallest possible.
G.2.3 Applications
Keywords and phrases unextendible product basis; quantum entanglement; graph factorization
Unextendible product bases play a rather diverse and important role in quantum informationtheory [7]. While their original motivation was for the construction of bound entangledstates [5, 12, 13], they have also been used to build indecomposible positive maps [14], todemonstrate Bell inequalities without a quantum violation [3], and demonstrate the existenceof nonlocality without entanglement [4].Furthermore, in the qubit case (i.e., the case where each local space has dimension 2),it has been shown that unextendible product bases can be used to construct tight Bellinequalities with no quantum violation [2] and subspaces of small dimension that are locallyindistinguishable [8]. It is the qubit case that we focus on in the present paper. In particular,we consider the question of how small a qubit unextendible product basis can be.The minimum cardinality of a qubit unextendible product basis on p qubits is well-knownto equal p + 1 when p is odd [1]. When p is even, however, the problem is more difficult. Itwas shown in [9] that the minimum cardinality equals p + 2 when p = 4 or p ≡ p ≥ p ≡ p + 3 when p = 8 and p + 4 in all othercases.Our approach is as follows: we formally introduce the mathematical preliminaries andgraph theory techniques that we make use of in Section 2. We construct unextendibleproduct bases of the claimed cardinality in Section 3. Finally, Section 4 is devoted to theproof that there does not exist a smaller unextendible product basis in these cases. A pure quantum state is represented by a unit vector | v i ∈ C d ⊗ · · · ⊗ C d p (and in oursetting, d = · · · = d p = 2 always). We say that | v i is a product state if we can write it in The Minimum Size of Qubit Unextendible Product Bases the form | v i = | v i ⊗ · · · ⊗ | v p i with | v j i ∈ C ∀ j. An unextendible product basis (UPB) is an orthonormal set S ⊆ ( C ) ⊗ p of productstates such that there is no product state orthogonal to every member of S . It is clear thatevery UPB in ( C ) ⊗ p contains at least p + 1 states – if it contained only p product states | v i , . . . , | v p − i then we could construct another product state that is, for each 0 ≤ j < p ,orthogonal to | v j i on the ( j + 1)-th party and thus violate unextendibility.It turns out that the trivial lower bound of p + 1 states can be attained when p is odd,and can almost be attained when p is even, as indicated by our main result: ◮ Theorem 1.
Let f ( p ) be the smallest possible number of states in a UPB in ( C ) ⊗ p . Then:(a) if p is odd then f ( p ) = p + 1 ;(b) if p = 4 or p ≡ mod then f ( p ) = p + 2 ;(c) if p = 8 then f ( p ) = p + 3 ;(d) otherwise, f ( p ) = p + 4 . Case (a) of Theorem 1 is demonstrated by the “GenShifts” UPB constructed in [7].Case (b) of Theorem 1 was proved in [9], and in general our techniques and presentationare similar to those of that paper. Our contribution is to prove cases (c) and (d) andhence complete the characterization. It is worth pointing out that cases (c) and (d) ofTheorem 1 are the first known cases (qubit or otherwise) where the minimum cardinality ofa UPB exceeds the trivial lower bound 1 + P j ( d j −
1) by more than 1 (see [6, 9] for severalexamples where the trivial lower bound is exceeded by exactly 1).Orthogonality graphs provide a very useful tool when dealing with unextendible productbases, particularly in the qubit case. Given a set of product states S = {| v i , . . . , | v s − i} ⊆ ( C ) ⊗ p with |S| = s , we say that the orthogonality graph of S is the graph on s vertices V := { v , . . . , v s − } such that there is an edge ( v i , v j ) of color ℓ if and only if | v i i and | v j i are orthogonal to each other on party ℓ . Rather than actually using p colors to color theedges of the orthogonality graph, for ease of visualization we instead draw p different graphson the same set of vertices – one for each party (see Figure 1). v v v v v v v v v v v v v v v v v v v v v Figure 1
The orthogonality graph of a set of 7 product states in ( C ) ⊗ . This set of states is aproduct basis, since every edge is present in at least one of the three graphs, but it is extendible,since we can find a product state that is orthogonal to the states associated with v , v , v , v onthe first subsystem, v , v on the second subsystem, and v on the third subsystem. The requirement that S is an orthonormal set is equivalent to requiring that every edgeis present on at least one party in its orthogonality graph. In order to help us visual-ize the unextendibility requirement, we make a few more observations. In particular, if . Johnston 3 | w i , | w i , | w i ∈ C are such that h w | w i = h w | w i = 0, then it is necessarily the casethat | w i = | w i (up to irrelevant complex phase). It follows that the orthogonality graphassociated with any qubit in a product basis is the disjoint union of complete bipartitegraphs. For example, in Figure 1 the left graph is K , , the center graph is the disjointunion of K , and K , , and the right graph is the disjoint union of K , and two copies of K , .Furthermore, not only does every set of product states have an orthogonality graphthat can be decomposed into the disjoint union complete bipartite graphs, but the converseis also true: every graph that is built from complete bipartite graphs in this way is theorthogonality graph of some set of product states. To see this, on each party assign to eachcomplete bipartite graph a distinct basis of C in the obvious way. For example, one set ofproduct states giving rise to the orthogonality graph depicted in Figure 1 is as follows: | v i := | i ⊗ | i ⊗ | i , | v i := | i ⊗ | i ⊗ | + i , | v i := | i ⊗ | i ⊗ | i , | v i := | i ⊗ | + i ⊗ |−i , | v i := | i ⊗ | + i ⊗ | + i , | v i := | i ⊗ |−i ⊗ | b i , | v i := | i ⊗ |−i ⊗ | b ⊥ i , where | + i := √ ( | i + | i ), |−i := √ ( | i − | i ), and {| b i , | b ⊥ i} is any orthonormal basis of C not equal to {| i , | i} or {| + i , |−i} .It is often useful to draw orthogonality graphs of sets of qubit product states in a formthat makes their decomposition in terms of complete bipartite graphs more transparent –we draw shaded regions indicating which vertices are equal to each other (up to complexphase) on the given party, and lines between shaded regions indicate that all states in oneof the regions are orthogonal to all states in the other region on that party (see Figure 2). v v v v v v v v v v v v v v v v v v v v v Figure 2
A representation of the same orthogonality graph as that of Figure 1. Vertices withinthe same shaded region represent states that are equal to each other on that party. Lines betweenshaded regions indicate that every state within one of the regions is orthogonal to every state withinthe other region.
It now becomes straightforward to see whether or not a product basis is unextendiblejust by looking at its orthogonality graph. A set of product states is unextendible if andonly if there is no way to choose one shaded region on each party such that every vertex v , v , . . . , v s − is contained within at least one of the shaded regions. For example, theset of product states described by Figure 2 is extendible because we can choose the shadedregion containing v , v , v , v on the first subsystem, v , v on the second subsystem, and v , v on the third subsystem.The following simple lemma shows that, in an orthogonality graph of a UPB, everyshaded region must be connected to exactly one other shaded region via an edge. The Minimum Size of Qubit Unextendible Product Bases ◮ Lemma 2. If S ⊆ ( C ) ⊗ p is a UPB, then for all | v i ∈ S and all integers ≤ j ≤ p thereis another product state | w i ∈ S such that | v i and | w i are orthogonal on the j -th subsystem. Proof.
Suppose that there exists 1 ≤ j ≤ p and | v i := | v (1) i ⊗ · · · ⊗ | v ( p ) i ∈ S such that | v i is not orthogonal to any other member of S on the j -th subsystem. Because S is a productbasis, | v i must be orthogonal to every member of S on the remaining p − | v ⊥ ( j ) i is orthogonal to | v ( j ) i then the product state | v (1) i ⊗ · · · | v ( j − i ⊗| v ⊥ ( j ) i ⊗ | v ( j +1) i ⊗ · · · ⊗ | v ( p ) i is orthogonal to every element of S , which shows that S isextendible. ◭ An obvious corollary of Lemma 2 is that, in the orthogonality graph of a UPB, everyparty must have an even number of distinct shaded regions – a fact that will be very usefulin Section 4.
Recall that our goal is to show that the smallest UPB in ( C ) ⊗ consists of 11 states andthe smallest UPB in ( C ) ⊗ k consists of 4 k + 4 states when k ≥
3. Our first step towardthis goal is to construct a UPB of the desired size in these cases. ◮ Lemma 3.
There exists a UPB in ( C ) ⊗ consisting of states. Proof.
The result follows simply from demonstrating an orthogonality graph on 11 verticesthat satisfies the product basis and unextendibility requirements described in Section 2.Such an orthogonality graph is provided in Figure 3.
Figure 3
Orthogonality graphs demonstrating that there exists an 11-state UPB in ( C ) ⊗ . Indeed, it is straightforward (albeit tedious) to check that the 8 graphs depicted inFigure 3 contain all 55 possible edges between 11 vertices, so the corresponding productstates are mutually orthogonal. Unextendibility follows from the (also straightforward buttedious) fact that there is no way to choose a shaded region containing 2 vertices on 3different parties without at least 2 of them containing the same vertex. ◭ We note that the UPB of Lemma 3 was found by a combination of computer search andtweaking by hand, and it does not seem to generalize to other values of p in any natural way.On the other hand, the UPBs that we now construct of cardinality 4 k + 4 are much “tidier”. . Johnston 5 ◮ Lemma 4. If k ≥ then there exists a UPB in ( C ) ⊗ k consisting of k + 4 states. Proof.
We begin by defining a family of k + 1 graphs B j,k := ( V, E j ) for 0 ≤ j ≤ k , each onthe same set of 4 k + 4 vertices V := { v i , w i , x i , y i , : 0 ≤ i ≤ k } . The set of edges E j in thegraph B j,k is defined as follows: E j := (cid:8) ( v i , x ( i + j )(mod ( k +1)) ) , ( v i , y ( i + j )(mod ( k +1)) ) , ( w i , x ( i + j )(mod ( k +1)) ) , ( w i , y ( i + j )(mod ( k +1)) ) : 0 ≤ i ≤ k (cid:9) . The three graphs B , , B , , and B , in the k = 2 case are depicted in Figure 4. It isclear that the graph obtained by taking the union of all edges in all sets B j,k for 0 ≤ j ≤ k is K k +2 , k +2 , the complete bipartite graph on two sets of 2 k + 2 vertices. y x w v y x w v y x w v y x w v y x w v y x w v y x w v y x w v y x w v Figure 4
The graphs B , (left), B , (center), and B , (right), used in the construction of aUPB of size 12 in ( C ) ⊗ . We now define three sets of states S ( j ) = {| v ( j ) i i , | w ( j ) i i , | x ( j ) i i , | y ( j ) i i : 0 ≤ i ≤ k } ⊆ C that have orthogonality graphs B j,k for 0 ≤ j ≤ {| b i i , | b ⊥ i i} k +1 i =0 be distinct orthonormal bases of C (i.e., h b i | b ⊥ i i = 0 for all i , but |h b i | b j i| , |h b i | b ⊥ j i| , |h b ⊥ i | b ⊥ j i| / ∈{ , } whenever i = j ). Then let | v ( j ) i i := | w ( j ) i i := | b i i and | x ( j ) i i := | y ( j ) i i := | b ⊥ ( i − j )(mod ( k +1)) i , for 0 ≤ j ≤
2, which clearly results in the desired orthogonality graphs. Furthermore, eachset S ( j ) has the property that any state | z i ∈ C can be orthogonal to at most two elementsof S ( j ) – a fact that we will use later when discussing unextendibility.For each of the remaining k − B j,k (3 ≤ j ≤ k ), we construct sets of productstates S (2 j − , j − = {| v (2 j − , j − i i , | w (2 j − , j − i i , | x (2 j − , j − i i , | y (2 j − , j − i i : 0 ≤ i ≤ k } ⊆ C ⊗ C that have orthogonality graphs B j,k for 3 ≤ j ≤ k . To this end, define | v (2 j − , j − i i := | b i i ⊗ | b i i| w (2 j − , j − i i := | b i +( k +1) i ⊗ | b i +( k +1) i| x (2 j − , j − i i := | b ⊥ ( i − j )(mod ( k +1)) i ⊗ | b ⊥ ( i − j )(mod ( k +1))+( k +1) i| y (2 j − , j − i i := | b ⊥ ( i − j )(mod ( k +1))+( k +1) i ⊗ | b ⊥ ( i − j )(mod ( k +1)) i , which results in the desired orthogonality graphs.We now turn our attention to the complement graph of K k +2 , k +2 , which is simply thedisjoint union of two disjoint copies of K k +2 , the complete graph on 2 k + 2 vertices. We The Minimum Size of Qubit Unextendible Product Bases v w v w v w x y x y x y Figure 5
The graph K that is the disjoint union of two copies of K . denote this graph by K k +2 , and it is depicted in the k = 2 case in Figure 5. The graph K k +2 will be the orthogonality graph of the remaining 4 k − (3 + 2( k − k + 1 parties.Our goal now is to define sets of states S ( j ) = {| v ( j ) i i , | w ( j ) i i , | x ( j ) i i , | y ( j ) i i : 0 ≤ i ≤ k } ⊆ C for 2 k − ≤ j ≤ k − K k +2 . To this end, we recall that it is well-known that K k +2 alwayshas a 1-factorization [10, Theorem 9.1], so K k +2 clearly has a 1-factorization as well(see Figure 6). This 1-factorization decomposes K k +2 into 2 k + 1 distinct 1-regular span-ning subgraphs, and any such graph is clearly the orthogonality graph of the set of states {| b i , | b ⊥ i , . . . , | b k +1 i , | b ⊥ k +1 i} ⊂ C (under an appropriate labelling of the vertices). v w v w v w x y x y x y v w v w v w x y x y x y v w v w v w x y x y x y v w v w v w x y x y x y v w v w v w x y x y x y Figure 6
A 1-factorization of K , which is useful for constructing a UPB of size 12 in ( C ) ⊗ . Since the union of the sets of edges present in all of the graphs considered so far is the . Johnston 7 complete graph K k +4 , we know that the states in the set S := k O j =1 | v ( j ) i i , k O j =1 | w ( j ) i i , k O j =1 | x ( j ) i i , k O j =1 | y ( j ) i i : 0 ≤ i ≤ k are mutually orthogonal. To see why this set is unextendible, recall that any non-zeroproduct state can be orthogonal to at most 2 states on each of the first 3 subsystems, andat most 1 state on each of the remaining 4 k − · · (4 k −
3) = 4 k + 3 of these productstates. Since no nonzero product state can be orthogonal to all 4 k + 4 members of S , it isunextendible, which completes the proof. ◭ We now turn our attention to the problem of proving that the UPBs constructed in Section 3are the smallest possible. Because the main result of [1] tells us that the minimum cardinalityof a UPB in ( C ) ⊗ k is at least 4 k + 2, we only have to prove that there is no UPB ofcardinality 4 k + 2 when k ≥ k + 3 when k ≥
3. While theproof that there is no UPB of cardinality 4 k + 2 is relatively straightforward, the proofthat there is no UPB of cardinality 4 k + 3 is more involved and consists of many cases andsub-cases. We make use of a C script to solve some of the messier cases, while we solve thesimpler cases by hand.For the entirety of this section, we make use of partial orthogonality graphs , which arethe same as orthogonality graphs, except perhaps with some conditions unspecified. Forexample, in Figure 7 the lack of lines indicating orthogonality between shaded regions doesnot signify that there are no regions orthogonal to each other, but rather that we just don’tcare which regions are orthogonal to each other. Similarly, in Figure 8 there are vertices thatare drawn outside of any shaded region. This is intended to mean that we don’t care whatthe shaded region involving that vertex looks like. In general, we only specify the pieces ofthe orthogonality graphs that are relevant for our proofs.It will be convenient for us to let P , . . . , P k denote the 4 k different parties. We alsolet M j denote the maximum number of vertices contained within a single shaded region onparty P j (which is equal to the maximum number of states in the UPB that are equal to eachother on party P j ), and let C n,j denote the number of distinct shaded regions containingexactly n vertices on party j (i.e., C n,j is the number of distinct group of exactly n states inthe UPB that are equal to each other on party P j ). For example, in Figure 2, if the graphscorrespond to parties P , P and P , then M = 4, M = M = 2, C , = 1, C , = 1, C , = 1, C , = 3, C , = 5, and C , = 1. ◮ Lemma 5.
There is no UPB in ( C ) ⊗ k of cardinality k + 2 when k ≥ . Proof.
Suppose for a contradiction that there exists a UPB of cardinality 4 k + 2 in ( C ) ⊗ k .If it were the case that M j ≥ j , then we could find a product state that isorthogonal to the 3 corresponding states on that party and to any 1 of the product stateson each of the remaining 4 k − k + 2 elements of the UPB, whichviolates unextendibility. Hence M j ≤ ≤ j ≤ k . We now split into two cases. Case 1:
There is at most one party P j with M j = 2.Between the 4 k parties, there must be a total of (4 k + 2)(4 k + 1) / k + 6 k + 1 edgesin their orthogonality graphs. The 4 k − P j must be the disjoint union The Minimum Size of Qubit Unextendible Product Bases of 2 k + 1 copies of K , , for a total of at most (4 k − k + 1) = 8 k + 2 k − P j then needs at least (8 k + 6 k + 1) − (8 k + 2 k −
1) = 4 k + 2 edges. It iseasily seen, however, that the largest number of edges that the orthogonality graph of party P j can have is obtained when it is the disjoint union of k copies of K , and one copy of K , , which results in only 4 k + 1 edges, which gives the desired contradiction. Case 2:
There are two (or more) parties P i = P j with M i = M j = 2.It is not difficult to see that C ,ℓ ∈ { , } for all ℓ or else either Lemma 2 or unextendibilityis violated. Furthermore, it is not difficult to see that the unique (up to repositioningvertices and parties) way to have C ,ℓ = 2 for 3 distinct values of ℓ is given in Figure 7,and there is no way to have C ,ℓ for a fourth value of ℓ without violating unextendibility. Asimple calculation reveals that the maximum number of edges that can be obtained from theorthogonality graphs of these 3 parties is (2 k + 3) + 2(2 k + 2) = 6 k + 7. The orthogonalitygraphs of the remaining 4 k − k + 1 copies of K , , so theyeach have 2 k + 1 edges. Thus the total number of edges among the orthogonality graphs ofall 4 k parties is at most (6 k + 7) + (4 k − k + 1) = 8 k + 4 k + 4. This quantity is smallerthan the 8 k + 6 k + 1 required edges when k ≥
2, which gives the desired contradiction.
Figure 7
Partial orthogonality graphs of three parties that each have two sets of two equal states,used in the proof of case 2 of Lemma 5. There is no way to add another pair of equal states on anyparty without violating unextendibility. ◭ Note that the hypothesis of Lemma 5 that k ≥ k + 4 k + 4 ≥ k + 6 k + 1 in case 2 of the proof of the lemma when k = 1, so it may bepossible to fit all of the required edges into the orthogonality graphs. Indeed, it was shownin [9] that a UPB consisting of 4 k + 2 states in ( C ) ⊗ k exists in the k = 1 case.We now turn our attention to proving that there is no UPB of cardinality 4 k + 3 when k ≥
3. The idea and techniques used in the proof of this statement are quite similar to the4 k + 2 case, but there are more cases to consider. ◮ Lemma 6.
There is no UPB in ( C ) ⊗ k of cardinality k + 3 when k ≥ . Proof.
Suppose for a contradiction that there exists a UPB of cardinality 4 k + 3 in ( C ) ⊗ k .If there exists 1 ≤ j ≤ p such that M j ≥
4, then we can find a product state that isorthogonal to at least 4 corresponding states on party P j and to 1 of the product stateson each of the remaining 4 k − k + 3 elements of the UPB, whichviolates unextendibility. Hence M j ≤ j . Furthermore, this same argument showsthat if there exists i ≥ i partiesso that together they contain at least i + 3 vertices, then unextendibility will be violated.Finally, note that since 4 k + 3 is odd, Lemma 2 implies that M j ≥ j .We now split into 4 cases, depending on the value of max j { C ,j } (i.e., the maximumnumber of sets of 3 equal states on any party). . Johnston 9 Case 1: max j { C ,j } ≥ M j ≥ j , it easily follows that we can find shaded regions on two partiesthat contain 3 + 2 = 5 distinct vertices, which contradicts unextendibility. Case 2: max j { C ,j } = 2.Suppose without loss of generality that party P is such that C , = 2. Unextendibilityimmediately implies that C ,j = 0 for j ≥
2. Since there are 4 k − P , which is odd, there must be a copy of K , on this party, as in Figure 8. Since v is connected to only one other state on party P , it must be connected to 2 states on eachof 2 other parties. These sets of 2 vertices must be disjoint and must each contain one of v , v , v and one of v , v , v . Thus parties P and P , without loss of generality, are as inFigure 8, which clearly implies extendibility and rules out this case. v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v Figure 8
The (essentially unique) partial orthogonality graphs of parties P (left), P (center)and P (right) in case 2 of Lemma 6. Such a product basis is necessarily extendible, as we can finda product state that is orthogonal to the states corresponding to v and v on party P , v and v on party P , v and v on party P , and one of the 4 k − k − Case 3: max j { C ,j } = 0.Since M j = 2 for all j , simple parity arguments show that C ,j ∈ { , , , . . . } for every j .We now split into two sub-cases, depending on the value of max j { C ,j } (i.e., the maximumnumber of sets of 2 equal states on any party). Case 3(a): max j { C ,j } ≥ P has C , = 5. We first argue that there must be at least oneother party P with C , ≥
3. To see this, suppose the contrary – suppose that C ,j = 1for all j ≥
2. Then each of these 4 k − k + 2 edges to theorthogonality graph, for a total of (4 k − k + 2) = 8 k + 6 k − P contributes no more than 4 k + 2 edges, for a total of 8 k + 10 k edges among all 4 k parties.However, the complete graph on 4 k + 3 vertices has (4 k + 3)(4 k + 2) / k + 10 k + 3 edges,so there are at least 3 pairs of non-orthogonal product states in our set, which contradictsthe assumption that we are working with a UPB.We now pick an arbitrary party P = P , P . Because C , ≥
1, we are now able tochoose one shaded region on each of parties P , P , P such that 6 vertices are containedwithin these regions, which shows that unextendibility is violated. To this end, we chooseany shaded region on party P that contains two vertices, then we pick any shaded region onparty P that is disjoint from the two vertices we chose on party P , and finally we chooseany shaded region on party P that is disjoint from all four of the previously-chosen vertices(see Figure 9). Case 3(b): max j { C ,j } ≤ v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v Figure 9
An example of a partial orthogonality graph in case 3(a) of Lemma 6. Such a productbasis is necessarily extendible, as we can choose the shaded region containing v and v on party P ,the disjoint shaded region (i.e., the one containing v and v ) on party P , and the disjoint shadedregion (i.e., the one containing v and v ) on party P , for a total of 6 vertices on 3 parties. than 4 distinct parties P j for which C ,j ≥ k +4edges in its orthogonality graph, and each of the remaining 4 k − k + 2edges on its orthogonality graph, for a total of at most 4(2 k +4)+(4 k − k +2) = 8 k +8 k +8edges. The complete graph on 4 k + 3 vertices has (4 k + 3)(4 k + 2) / k + 10 k + 3 edges,so when k ≥ k = 2 case, sothe fact that we require k ≥ Case 4: max j { C ,j } = 1.By parity arguments, we see that every party P j with C ,j = 1 must also have C ,j ∈{ , , , . . . } . Furthermore, if there exist two (or more) parties P , P such that M = M = 3,then unextendibility is violated unless C ,j = 1 whenever M j = 3. Case 4(a):
There exist three (or more) parties P , P , P such that M = M = M = 3.Because there must exist a shaded region containing exactly 2 vertices on each party P , P , P , it is easily verified that the only possible configuration of shaded regions on thoseparties (up to repositioning vertices and parties) that doesn’t break unextendibility is theone depicted in Figure 10. Figure 10
The (essentially unique) partial orthogonality graph that does not violate unextend-ibility in case 4(a).
The parties P , P , P can have no more than (2 k + 5)+ 2(2 k + 3) = 6 k + 11 distinct edgesamong them (since there will be a lot of overlap at the left edge of the graphs if we make eachgroup of 3 equal states orthogonal to the group of 2 equal states). It is straightforward to seethat none of the remaining 4 k − P j can have M j ≥ C ,j ≥ k − k + 2 edges each, for atotal of 6 k +11+(4 k − k +2) = 8 k +8 k +5 edges. Since 8 k +8 k +5 < k +10 k +3 when . Johnston 11 k ≥
2, there are some edges missing from the orthogonality graphs, which is a contradiction.
Case 4(b):
There exists a party P such that M = 3, but M j ≤ j ≥ P contributes at most 2 k + 5 edges to the orthogonality graph, and the unex-tendibility requirement implies that C ,j ≤ j ≥
2. Suppose that there are m indices2 ≤ j , j , . . . , j m ≤ k such that C ,j i = 3 for 1 ≤ i ≤ m and C ,j = 1 for all other values of j . Then there are at most (2 k + 5) + m (2 k + 4) + (4 k − m − k + 2) = 8 k + 8 k + 2 m + 3total edges between all 4 k parties. As in the previous cases, we need a total of 8 k + 10 k + 3edges, which implies that m ≥ k . We already saw via brute-force search in case 3(b) thatwe can’t have m ≥
5, so we only need to rule out the 3 ≤ k ≤ P is represented by vertices v , v , and v (see Figure 11), then each one of the 3 groups of 2 identical states on the other parties mustcontain exactly one of v , v , or v . By refining our brute-force computer search to take thisrestriction into account, we find that there is no configuration of shaded regions that doesnot violate unextendibility when m ≥ k ≥ v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v Figure 11
An example of a partial orthogonality graph in case 4(b).
Case 4(c):
There exist two parties P , P such that M = M = 3, but M j ≤ j ≥ P and P , which are depicted in Figures 12 and 13. Notice that inFigure 12, the shaded region on party P that contains exactly two vertices does not shareany common vertices with the shaded region on party P that contains exactly two vertices,while in Figure 13 those two regions contain the common vertex v . v v v v v v v v v v v v v v v v v v Figure 12
One of two possible partial orthogonality graphs of parties P , P , and P that doesnot violate unextendibility in case 4(c). Suppose for now that parties P and P have a total of at most 4 k + 8 distinct edges ontheir orthogonality graphs. If there are m parties P j ( j ≥
3) for which C ,j = 3, then wehave a total of at most (4 k + 8) + m (2 k + 4) + (4 k − m − k + 2) = 8 k + 8 k + 2 m + 4edges. Any all of these m parties, we require that one of the shaded regions contains v and v v v v v v v v v v v v Figure 13
The other possible partial orthogonality graph of parties P , P that does not violateunextendibility in case 4(c). v and the other shaded regions containing two vertices each contain one of v or v . Thus,the brute-force search described in case 4(b) applies here as well and shows that m ≤ m = 2 we have 8 k + 8 k + 2 m + 4 = 8 k + 8 k + 8 < k + 10 k + 3 when k ≥
3, which shows that there can not possibly be enough edges on the orthogonality graphsin this case.The only remaining possibility is that the parties P and P have a total of at least4 k + 9 distinct edges (and hence exactly k + 9 distinct edges). In this case, parties P and P must be as in Figure 12, and on both of the parties P and P the set of 3 equal statesmust be orthogonal to the set of 2 equal states. Furthermore, it is not difficult to show thatin this case, any party P j with C ,j = 3 can introduce at most 2 k + 3 new edges that arenot already present in the orthogonality graph of parties P and P . Thus, if there are m parties P j ( j ≥
3) for which C ,j = 3, we have a total of at most (4 k + 9) + m (2 k + 3) +(4 k − m − k + 2) = 8 k + 8 k + m + 5 edges. Since m ≤ k ≥
3, it followsthat 8 k + 8 k + m + 5 < k + 10 k + 3, which again shows that there can not possibly beenough edges on the orthogonality graphs in this case. ◭ Acknowledgements
Thanks are extended to Gus Gutoski for suggesting a computer searchto fill in the gaps in the proof of Lemma 6. The author was supported by the Natural Sciencesand Engineering Research Council of Canada and the Mprime Network.
References N. Alon and L. Lovász. Unextendible product bases.
J. Combinatorial Theory, Ser. A ,95:169–179, 2001. R. Augusiak, T. Fritz, M. Kotowski, M. Kotowski, M. Pawłowski, M. Lewenstein, andA. Acín. Tight Bell inequalities with no quantum violation from qubit unextendible productbases.
Phys. Rev. A , 85:042113, 2012. R. Augusiak, J. Stasinska, C. Hadley, J. K. Korbicz, M. Lewenstein, and A. Acín. Bellinequalities with no quantum violation and unextendible product bases.
Phys. Rev. Lett. ,107:070401, 2011. C. H. Bennett, D. P. DiVincenzo, C. A. Fuchs, T. Mor, E. Rains, P. W. Shor, J. A. Smolin,and W. K. Wootters. Quantum nonlocality without entanglement.
Phys. Rev. A , 59:1070–1091, 1999. C. H. Bennett, D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal.Unextendible product bases and bound entanglement.
Phys. Rev. Lett. , 82:5385–5388, 1999. J. Chen and N. Johnston. The minimum size of unextendible product bases in the bipartitecase (and some multipartite cases). E-print: arXiv:1301.1406 [quant-ph], 2013. . Johnston 13 D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal. Unextendibleproduct bases, uncompletable product bases and bound entanglement.
Commun. Math.Phys. , 238:379–410, 2003. R. Duan, Y. Xin, and M. Ying. Locally indistinguishable subspaces spanned by three-qubitunextendible product bases.
Phys. Rev. A , 81:032329, 2010. K. Feng. Unextendible product bases and 1-factorization of complete graphs.
Discrete Appl.Math. , 154:942–949, 2006. F. Harary.
Graph Theory . Addison-Wesley, Reading, Mass., 1969. N. Johnston. Code for proving that no UPB of size 4 k + 3 exists on 4 k qubits. Publishedelectronically at , 2013. J. M. Leinaas, P. Ø. Sollid, and J. Myrheim. Unextendible product bases and extremaldensity matrices with positive partial transpose. E-print: arXiv:1104.1318 [quant-ph], 2011. Ł. Skowronek. Three-by-three bound entanglement with general unextendible productbases.
J. Math. Phys. , 52:122202, 2011. B. M. Terhal. A family of indecomposable positive linear maps based on entangled quantumstates.