aa r X i v : . [ qu a n t - ph ] N ov Absolutely entangled set of pure states
Mao-Sheng Li
1, 2, ∗ and Man-Hong Yung
1, 3, † Department of Physics, Southern University of Science and Technology, Shenzhen, 518055, China Department of Physics, University of Science and Technology of China, Hefei, 230026, China Institute for Quantum Science and Engineering, and Department of Physics,Southern University of Science and Technology, Shenzhen, 518055, China
Quite recently, Cai et al. [arXiv:2006.07165v1] proposed a new concept “absolutely entangledset” for bipartite quantum systems: for any possible choice of global basis, at least one state of theset is entangled. There they presented a minimum example with a set of four states in two qubitsystems and they proposed a quantitative measure for the absolute set entanglement. In this work,we derive two necessity conditions for a set of states to be an absolutely entangled set. In addition,we give a series constructions of absolutely entangled bases on C d ⊗ C d for any nonprime dimension d = d × d . Moreover, based on the structure of the orthogonal product basis in C ⊗ C n , we obtainanother construction of absolutely entangled set with 2 n + 1 elements in C ⊗ C n . I. INTRODUCTION
The phenomenon of entanglement [1] has been identi-fied as an important resource in many quantum informa-tion processing such as quantum teleportation [2, 3] andquantum key distribution[4, 5]. Entanglement is a con-cept on compounded quantum systems. So its existenceis relative to the partition of the subsystems. However,the way to subdivide a given physical system S into sub-systems may not be unique in general. Sometimes, thepartition of subsystems depend on the physical resourcesavailable to access and manipulate the degrees of free-dom of a quantum system [6, 7]. There are some stateswhich is entangled with respect to some subsystems butseparable with respect to another subsystems.The concept of absolutely entangled set was proposedquite recently. A set of bipartite states is called an ab-solutely entangled set if there always exist some statesin the entangled form after any action of global unitary.The study of the structure of absolutely entangled setmay be helpful as operational way to the device indepen-dent certification of entangled measurement [8]. As a newconcept, there are still many problems worthy of furtherstudy. Given a set of states, how to identify whether it isabsolutely entangled or not? How to construct small setof absolutely entangled set for general dimensional sys-tem? In this article, we will give partial answers to thesequestions.The rest of this article is organized as follows. In Sec.II, we give some necessary notation, and definition ofthe absolutely entangled set. In Sec. III, we study twoproperties on the existence of absolutely entangled set. InSec. IV, we give two concrete constructions of absolutelyentangled sets. Finally, we draw a conclusion and presentsome interesting problems in Sec. V. ∗ [email protected] † [email protected] II. NOTATION AND DEFINITION OFABSOLUTELY ENTANGLED SET
Throughout this paper, we use the following notation.For a give positive integer N , denote [ N ] to be the set { , · · · , N } . Given a nonprime integer d with d = d × d where both d , d ≥ C d and C d ⊗ C d , i.e., C d = C d ⊗ C d .And we use U( d ) to denote the set of all unitary matricesof dimensional d . The states we focus here are mostlypure states, but sometimes without normalization. Definition 1
Consider a set of pure quantum states {| ψ i , ..., | ψ n i} in a fixed Hilbert space C d of non-primedimension. The set is said to be absolutely entangled withrespect to all bipartitions into subsystems of dimension ( d , d ) , if for every unitary U ∈ U( d ) , at least one state U | ψ j i is entangled with respect to C d ⊗ C d . III. NECESSITY CONDITION FORABSOLUTELY ENTANGLED SET
Given a set S of states in C d ⊗ C d , we would liketo map the states in S to a set of product states by aunitary matrix. The most simple set of product states isthe form | i⊗ C d ⊂ C d ⊗ C d . Therefore, it is interestingto ask under what condition can we embed S to the set | i ⊗ C d . In fact, the basic linear algebra course gives usthat S can be embedded to the set | i ⊗ C d if and onlyif dim C ( S ) ≤ d . Therefore, a necessity condition for S to be an absolutely entangled set is dim C ( S ) ≥ d + 1 . Proposition 1
Let d ≥ be a non-prime positive integerand suppose d = d d with ≤ d ≤ d . Suppose that S ≡ {| ψ j i} Nj =1 ⊆ C d is an absolutely entangled set withrespect to ( d , d ) . Then we have dim C ( S i ) ≥ d +1 where S i ≡ S \ {| ψ i i} for each i ∈ [ N ] .Proof . Suppose not, there exists some i such thatdim C ( S i ) ≤ d . If dim C ( S i ) < d , then dim C ( S ) ≤ d .However, in this case, S be embedded in to the sub-space | i ⊗ C d which is contradicted with that S isan absolutely entangled set. So we can assume thatdim C ( S i ) = d . So there exists a set of d orthogonalstates B ≡ {| ξ k i} d k =1 ⊆ C d such thatspan C ( S i ) = span C ( B ) . For each | ψ j i ∈ S i , the state | ψ j i can be written as linearcombination of {| ξ k i} d k =1 , i.e. | ψ j i = P d k =1 c jk | ξ k i . Anystate | ψ i ∈ C d can be written as the form | ψ i = α | ξ i + β | η i where | ξ i ∈ span C ( S i ) and | η i ∈ span C ( S i ) ⊥ (theorthogonal complementary space of span C ( S i )). Spe-cially, we can suppose that | ψ i i = α | ξ i + β | η i . As | ξ i ∈ span C ( S i ), the state | ψ i i can be written as | ψ i i = ( d X k =1 c ik | ξ k i ) + β | η i . (i). If c ik = 0 for all k , then | ψ i i = β | η i . We can definea unitary U ∈ U( d ) with the following conditions U | ξ k i = | i| k i , k = 1 , · · · , d , and U | η i = | i| i .Obviously, under this map, the states in S are beingsent to product states.(ii). If c ik = 0 for some k , we can define a unitary U in U( d ) with the following conditions U | ξ k i = | i| k i , k = 1 , · · · , d , and U | η i = | i ( N [ d X k =1 c ik | k i ])where N is a normalized factor. Under this map,we have U | ψ j i = | i ( P d k =1 c jk | k i ) for j = i and U | ψ i i = | i ( d X k =1 c ik | k i ) + c id +1 | i ( N [ d X k =1 c ik | k i ])= ( | i + N c id +1 | i )( d X k =1 c ik | k i ) . Both cases are contradicted with the assumption that S is an absolutely entangled set.The above result is more general than the one pro-posed in Ref. [10]. In the following, we present anotherproperty on the existence of the absolutely entangled setwhich may not cover by the above necessity. Proposition 2
Let d ≥ be a non-prime positive in-teger and suppose d = d d with ≤ d ≤ d . Let S ≡ {| ψ j i} Nj =1 ⊆ C d be a set of pure states. If S can be seperated into k disjoint sets S i ≡ {| ψ ( i ) j i} N i j =1 (here k ≤ d and P ki =1 N i = N ) such that the statescome from different sets are orthogonal to each other and dim C ( S i ) ≤ d , then there exists some unitary U ∈ U( d ) such that { U | ψ j i} Nj =1 are all in product form.Proof . Let {| s i| t i| ≤ s ≤ d , ≤ t ≤ d } denote acomputational basis of the bipartite system H A ⊗ H B = C d ⊗ C d . Denote V i ≡ span C ( S i ) for i = 1 , · · · , k and V ≡ span C ( S ). By the orthogonality conditions, we canconclude that there is a direct sum decomposition V = k M i =1 V i . Denote n i ≡ dim C ( V i ) = dim C ( S i ). There exists an or-thonormal basis B i ≡ {| ξ ( i ) j i} n i j =1 of V i for each i . Bythe direct sum decomposition, we have B i ∩ B j = ∅ if1 ≤ i = j ≤ k and ∪ ki =1 B i is an orthonormal basis of V .So ∪ ki =1 B i can be extended to be a complete orthonormalbasis of the whole space C d . Therefore, there exists someunitary U ∈ U( d ) who maps the basis B to the compu-tational basis {| s i| t i| ≤ s ≤ d , ≤ t ≤ d } . As n i ≤ d and k ≤ d , we can also assume that U | ξ ( i ) j i = | i i| j i , i = 1 , · · · , k, and j = 1 , · · · , n i . By the definitions of V i and B i , each | ψ ( i ) l i ∈ S i can bewritten as a linear combination of {| ξ ( i ) j i} n i j =1 . That is, | ψ ( i ) l i = n i X j =1 c ( i ) lj | ξ ( i ) j i , where c ( i ) lj ∈ C and P n i j =1 | c ( i ) lj | = 1. Then by the defi-nition of the unitary U , we have U | ψ ( i ) l i = n i X j =1 c ( i ) lj | i i| j i = | i i ( n i X j =1 c ( i ) lj | j i )which is a product state in the bipartite system H A ⊗H B . IV. CONSTRUCTIONS OF ABSOLUTELYENTANGLED SETS
In Ref. [10], the authors presented a absolutely en-tangled set with four elements in C = C ⊗ C . Moreexactly, they proved that the set consisting of the follow-ing four states is an absolutely entangled set | ψ i = | ξ i , | ψ i i = c i | ξ i + c ii | ξ i i , i = 2 , , {| ξ i i} i =1 is an orthonormal basis of C and | c i | ∈ ( ,
1) for i = 2 , ,
4. Here we give an general constructionof absolutely entangled set for high dimensional systems.
Theorem 1
Let d ≥ be a non-prime positive integerand | ξ i , · · · , | ξ d i be an orthonormal basis of C d . Thenthe set consisting of the following d states | φ i ≡ | ξ i ; | φ i i ≡ a i | ξ i + b i | ξ i i ; i = 2 , · · · , d forms an absolutely entangled set provided the conditions λ | a i | ≥ | b i | / ( i = 2 , · · · , d ) for some λ ∈ (0 , . Proof . Suppose not, there exists some unitary matrix U ∈ U( d ) such that { U | ψ i i} di =1 are all product states in C d ⊗ C d (here we assume d = d d ). Without loss ofgenerality, we assume that U | ξ i = | i| i . As any unitarymatrix preserves orthogonal relations, we have U | ξ i i = X ( k,l ) =(1 , u i,c kl | k i| l i . Here the ( u i,c kl ) is a ( d − × ( d −
1) unitary matrix. Bydefinition, U | ψ i i = a i | i| i + b i X ( k,l ) =(1 , u i,c kl | k i| l i , for i = 2 , · · · , d are all in product form. Therefore, each of the followingmatrix a i b i u i,c b i u i,c · · · b i u i,c d b i u i,c b i u i,c b i u i,c · · · b i u i,c d b i u i,c b i u i,c b i u i,c · · · b i u i,c d · · · ... ... · · · ... b i u i,c d b i u i,c d b i u i,c d · · · b i u i,c d d is rank one. Particularly, the left top 2 × a i b i u i,c − b i u i,c b i u i,c = 0 whichis equivalent to a i u i,c = b i u i,c u i,c as b i = 0. ByCauchy inequality, | a i |·| u i,c | = | b i |·| u i,c |·| u i,c | ≤ | b i | | u i,c | + | u i,c | . Squaring both sides of the above inequality, we obtain | a i | · | u i,c | ≤ | b i | ( | u i,c | + | u i,c | ) ≤ | b i | ( | u i,c | + | u i,c | ) . The second inequality holds because of | u i,c | + | u i,c | ≤
1. As λ | a i | > | b i | / | u i,c | ≤ λ ( | u i,c | + | u i,c | ) / i = 2 , · · · , d . Summing up the two sides of i from 2to d , we obtain1 = d X i =2 | u i,c | ≤ λ ( d X i =2 | u i,c | + d X i =2 | u i,c | ) / λ < . Wherehence, we obtain a contradiction. So the originalset must be absolutely entangled.In the following, we will present another example ofabsolutely entangled set in C . However, the number ofelements of our set here is five instead of four. Example 1
Let d = 4 , p = 2 , and {| ξ i i} i =1 be an or-thonormal basis of C . Then the set consisting of thefollowing five states | ψ i i = | ξ i i , i = 1 , , , | ψ i = X j =1 x p j | ξ j i , | ψ i = X j =1 x p j | ξ j i is an absolutely entangled set for almost all x ∈ (0 , (That is, there are only finitely exceptional cases).Proof. Suppose not, there is a unitary transformation U ∈ U(4) such that { U | ψ i i} i =1 are all in product formof H A ⊗ H B = C ⊗ C . Particularly, { U | ξ i i} i =1 are or-thonormal product states in H A ⊗ H B . As any unitarypreserves the orthogonality, U | ξ i should be also orthog-onal to all { U | ξ i i} i =1 . As it is well known that there is noUPB whose complementary space is of dimensional one.That is, any set of orthogonal product states with d − { U | ξ i i} i =1 up to a global phase. Hence U | ξ i must alsobe in a product form. So the unitary U maps the basis {| ξ i i} i =1 to an orthonormal product basis of H A ⊗ H B .It is not difficult to show that the general orthonormalproduct basis of H A ⊗ H B is of the form (or an exchangethe two particles) | α i| β i , | α ⊥ i| γ i , | α i| β ⊥ i , | α ⊥ i| γ ⊥ i . Then U | ψ i and U | ψ i can be written as | α i ( x p i | β i + x p j | β ⊥ i ) + | α ⊥ i ( x p k | γ i + x p l | γ ⊥ i ) , | α i ( x p i | β i + x p j | β ⊥ i ) + | α ⊥ i ( x p k | γ i + x p l | γ ⊥ i ) . respectively. One necessary condition for the above twostates to be in product form is that x p i | β i + x p j | β ⊥ i //x p k | γ i + x p l | γ ⊥ i ; x p i | β i + x p j | β ⊥ i //x p k | γ i + x p l | γ ⊥ i . (1)Here i, j, k, l is a complete list of the four numbers1 , , ,
4. Suppose that | γ i = v | β i + v | β ⊥ i , | γ ⊥ i = v | β i + v | β ⊥ i . Then the matrix V ≡ ( v ij ) is a unitary matrix. Therelations in Eq. (1) are equivalent to that N ([ x p k , x p l ]) V = e iθ N ([ x p i , x p j ])and N ([ x p k , x p l ]) V = e iθ N ([ x p i , x p j ]) . Here and below we use the notation N ( v ) to denote thevector v k v k for any nonzero vector v .Noticing that any unitary matrix should preserve theinner product of vectors. As x is a real number by as-sumption, the following two terms x p k + x p l p x p k + x p l p x p k + x p l , and x p i + x p i p x p i + x p j p x p i + x p j are equal. This is equivalent to the equation ( x p k + x p l ) ( x p k + x p l )( x p k + x p l ) = ( x p i + x p j ) ( x p i + x p j )( x p i + x p j ) . (2) We define f ( p ) ij | kl ( X ) as the following f ( p ) ij | kl ( X ) ≡ ( X p k + X p l ) ( X p i + X p j )( X p i + X p j ) − ( X p i + X p j ) ( X p k + X p l )( X p k + X p l ) . Using mathematica, we can calculate the three expres-sions f (2)12 | ( X ), f (2)13 | ( X ) , and f (2)14 | ( X ) respectively asfollows: − X + 2 X − X + X + 2 X − X − X − X − X + 2 X + X − X + 2 X − X , − X + 2 X − X + 2 X − X − X + 2 X − X + 2 X − X ,X − X + X − X − X + 2 X + 2 X − X − X + 2 X + 2 X − X − X + X − X + X . TABLE I. The real roots of the above three polynomials for constructing absolutely entangled set in C .PolyRoots Num 1 2 3 4 5 6 7 f (2)12 | ( X ) -1.21341 -1 -0.824127 0 0.824127 1 1.21341 f (2)13 | ( X ) -1.10104 -1 -0.908231 0 0.908231 1 1.10104 f (2)14 | ( X ) -1.11046 -1 -0.900525 0 0.900525 1 1.11046 There are only fifteen real numbers x i (see Table I) s. t. f (2)12 | ( x i ) f (2)13 | ( x i ) f (2)14 | ( x i ) = 0 . So for any real number x = x i for all i = 1 , · · · ,
15, the setconsisting of the five states we constructed above indeedforms an absolutely entangled set.In fact, we idea of Example 1 can be extended to thebipartite system C ⊗ C n . But we need some more re-sults on the structure of product basis in the system of C ⊗ C n . Based on the structure results, we will presentanother construction of absolutely entangled set of theform (2 , n ). Now we present the statement of some knownresults about orthnormal product basis in C ⊗ C n . Lemma 1 (Feng 2006, see Ref. [11])
Let H A = C and H B = C n . Any orthonormal product basis of H A ⊗H B can be written in the following way: S = {| a k i ⊗ A k , | a ⊥ k i ⊗ A ′ k (cid:12)(cid:12) ≤ k ≤ t } where {| a k i , | a ⊥ k i} are different (normalized) orthogonalbasis in H A and A k = {| u ( k ) i i} n k i =1 , A ′ k = {| v ( k ) i i} n ′ k i =1 ⊆H B ( ≤ k ≤ t ) such that the following conditions hold(1) h u ( k ) i | u ( k ) j i = δ ij ;(2) h v ( k ) i | v ( k ) j i = δ ij ;(3) n k = span C ( A k ) = span C ( A ′ k ) = n ′ k ; (4) H B = L tk =1 span C ( A k ) . Theorem 2
Let d ≥ be an even integer and suppose d = 2 n . Let {| ξ i i} di =1 be an orthonormal basis of C d .Then the set consisting of the following d + 1 states | ψ i i = | ξ i i , i = 1 , , · · · , d − , | ψ d i = d X j =1 x p j | ξ j i , | ψ d +1 i = d X j =1 x p j | ξ j i is an absolutely entangled set of the form (2 , n ) for al-most all x ∈ (0 , provided p to be an integer greaterthan 7.Proof. Suppose not, there is a unitary transformation U ∈ U( d ) such that { U | ψ i i} d +1 i =1 are all in product formof H A ⊗ H B = C ⊗ C n . Particularly, { U | ξ i i} d − i =1 areorthonormal product states in H A ⊗ H B . As any unitarypreserves the orthogonality, U | ξ d i should be orthogonalto all { U | ξ i i} d − i =1 . Similar with the argument of Exam-ple 1, the unitary U maps the basis {| ξ i i} di =1 to an or-thonormal product basis of H A ⊗ H B . By Lemma 1, anyorthonormal product basis of H A ⊗ H B is of the form S = {| a k i ⊗ A k , | a ⊥ k i ⊗ A ′ k (cid:12)(cid:12) ≤ k ≤ t } where | a i i , | a ⊥ i i , A i , A ′ i satisfy the four conditions in thestatement of Lemma 1.Then U | ψ d i and U | ψ d +1 i can be written as U | ψ d i = t X k =1 ( n k X i =1 x p gki | a k i| u ( k ) i i + n k X i =1 x p hki | a ⊥ k i| v ( k ) i i ) ,U | ψ d +1 i = t X k =1 ( n k X i =1 x p gki | a k i| u ( k ) i i + n k X i =1 x p hki | a ⊥ k i| v ( k ) i i ) . (3) Here g ki , h ki , ( k = 1 , · · · , t, i = 1 , · · · , n k ) is a completelist of 1 , , , · · · , d . We separate the analysis into twocases: (1) all n k are equal to 1; (2) at least one n k ≥ t = n , | u ( k )1 i = | v ( k )1 i up to a phaseand {| u ( k )1 i} nk =1 form an orthonormal basis of H B . Eqs.(3) can be simplified as U | ψ d i = n X k =1 ( x p gk | a k i + x p hk | a ⊥ k i ) | u ( k )1 i ,U | ψ d +1 i = n X k =1 ( x p gk | a k i + x p hk | a ⊥ k i ) | u ( k )1 i . (4)One necessary condition for the two states in Eqs. (4) tobe in product form is that for k = tx p gk | a k i + x p hk | a ⊥ k i //x p gt | a t i + x p ht | a ⊥ t i ; x p g k | a k i + x p hk | a ⊥ k i //x p gt | a t i + x p ht | a ⊥ t i . As g k , h k , g t , h t are four different elements in { , , · · · , d } , we use i, j, k, l to denote this four elementsrespectively for simplicity. Under this simplification, theabove relations are equivalent to that there exists an 2 × V of such that N ([ x p k , x p l ]) V = e iθ N ([ x p i , x p j ])and N ([ x p k , x p l ]) V = e iθ N ([ x p i , x p j ]) . And the unitary matrix should preserve the inner productof vectors. As x is a real number by assumption, thefollowing two terms x p k + x p l p x p k + x p l p x p k + x p l , and x p i + x p i p x p i + x p j p x p i + x p j are equal. This is equivalent to the equation( x p k + x p l ) ( x p k + x p l )( x p k + x p l ) = ( x p i + x p j ) ( x p i + x p j )( x p i + x p j ) . (5)We define f ( p ) ij | kl ( X ) as the following f ( p ) ij | kl ( X ) ≡ ( X p k + X p l ) ( X p i + X p j )( X p i + X p j ) − ( X p i + X p j ) ( X p k + X p l )( X p k + X p l ) . Therefore, in this case, the x should be a root of f ( p ) ij | kl ( X ) = 0 by Eq. (5).(2) Suppose that n k ≥ k . One necessarycondition for the two states in Eqs. (3) to be in productform is that n k X i =1 x p gki | u ( k ) i i // n k X i =1 x p hki | v ( k ) i i ; n k X i =1 x p gki | u ( k ) i i // n k X i =1 x p hki | v ( k ) i i . This is equivalent to that there exists an n k × n k unitarymatrix W such that N ([ x p hk , · · · , x p hknk ]) W = e iφ N ([ x p gk , · · · , x p gknk ])and N ([ x p hk , · · · , x p hknk ]) W = e iφ N ([ x p gk , · · · , x p gknk ]) . And the unitary matrix W should preserve the innerproduct of the two vectors. That is, n k X i =1 x p hki vuut n k X i =1 x p hki vuut n k X i =1 x p hki , and n k X i =1 x p gki vuut n k X i =1 x p gki vuut n k X i =1 x p gki are equal. This is equivalent to the following equation ( n k X i =1 x p hki ) n k X i =1 x p hki n k X i =1 x p hki = ( n k X i =1 x p gki ) n k X i =1 x p gki n k X i =1 x p gki . (6) We define f ( p ) h k ...h knk | g k ...g knk ( X ) as the following f ( p ) h k ...h knk | g k ...g knk ( X ) = ( n k X i =1 X p hki ) n k X i =1 X p gki n k X i =1 X p gki − ( n k X i =1 X p gki ) n k X i =1 X p hki n k X i =1 X p hki . By Eq. (6), x should be a root of the equation f ( p ) h k ...h knk | g k ...g knk ( X ) = 0 . For each integer s (2 ≤ s ≤ n ), define P s to be the set { f ( p ) i ...i s | j ...j s ( X ) (cid:12)(cid:12) i , . . . , i s , j , . . . , j s ∈ [2 n ] and are different } where f ( p ) i ...i s | j ...j s ( X ) is defined as the Lemma 2. Andwe denote P := S ns =2 P s . To make all the states in the set we construct to beseparable, the x must be a root of some polynomial in P .Now set X := { x ∈ (0 , |∃ f ∈ P , s.t. f ( x ) = 0 } . However, there are only a finite number of polynomialsin P and each is nonzero by Lemma 2. Therefore, X is a finite set and any x ∈ (0 , \ X do not satisfy theEqs. (5) and (6). For such x , the set of d + 1 states weconstructed is an absolutely entangled set. Lemma 2
Let s ≥ and p ≥ be integers. Given any s different positive integers h , · · · , h s , g , · · · , g s , we definea polynomial f ( p ) h ...h s | g ...g s ( X ) as the following ( s X i =1 X p hi ) s X k =1 X p gk s X l =1 X p gl − ( s X k =1 X p gk ) s X i =1 X p hi s X j =1 X p hj . Then f ( p ) h ...h s | g ...g s ( X ) is a nonzero polynomial. Sketch of proof.
The first term and second term of f ( p ) h ...h s | g ...g s ( X ) can be expanded as s X i =1 s X j =1 s X k =1 s X l =1 X p hi +3 p hj +2 p gk +4 p gl , s X i ′ =1 s X j ′ =1 s X k ′ =1 s X l ′ =1 X p hi ′ +4 p hj ′ +3 p gk ′ +3 p gl ′ (7)respectively. As p ≥ h , · · · , h s , g , · · · , g s , the following equality holds3 p h i + 3 p h j + 2 p g k + 4 p g l = 2 p h i ′ + 4 p h j ′ + 3 p g k ′ + 3 p g l ′ if and only if i = j = i ′ = j ′ and k = l = k ′ = l ′ .Therefore, only s terms in the first expansion of Eq. (7)can be eliminated from the terms in the second expan-sion. From this, we can conclude that f ( p ) h ...h s | g ...g s ( X )is indeed a nonzero polynomial. V. CONCLUSION AND DISCUSSION
We have derived two necessary conditions for a set tobe an absolutely entangled set. This gives an improve-ment of that obtained in the original paper of absolutely entangled set [10]. We also presented two constructionsof the absolutely entangled sets on the systems C m ⊗ C n and C ⊗ C n respectively. The first construction is in-spired by the construction of Cai et al.’s results and itholds for general dimensional systems. The second oneis more technical but only holds in special dimensionalsystems.As there is no orthogonal product basis structure forgeneral bipartite systems, we do not know whether oursecond construction can be extended to more general di-mensional cases. And we do not consider the optimalconstruction in this paper. However, it is much moreinteresting to find out the smallest sets of absolutely en-tangled set of a given dimension. Acknowledgments
This work is supported byNational Natural Science Foundation of China(11875160, 12005092 and U1801661), the ChinaPostdoctoral Science Foundation (2020M681996), theNatural Science Foundation of Guang-dong Province(2017B030308003), the Key R&D Program of Guang-dong province (2018B030326001), the Guang-dongInnovative and Entrepreneurial Research TeamPro-gram (2016ZT06D348), the Science, Technology andInnovation Commission of Shenzhen Municipality(JCYJ20170412152620376 and JCYJ20170817105046702and KYTDPT20181011104202253), the Economy, Tradeand Information Commission of Shenzhen Municipality(201901161512). [1] R. Horodecki, P. Horodecki, M. Horodeckiand K. Horodecki. Quantum entanglement.Rev. Mod. Phys. , 865 (2009).[2] C. H. Bennett, G. Brassard, C. Cr´epeau, R. Jozsa, A.Peres, and W. K. Wootters. Teleporting an unknownquantum state via dual classical and Einstein-Podolsky-Rosen channels. Phys. Rev. Lett. , 1895 (1993).[3] D. Bouwmeester, J.-W. Pan, K. Mattle, M. Eibl, H. We-infurter and A. Zeilinger. Experimental quantum telepor-tation. Nature. , 575-579 (1997).[4] C. H. Bennett. Quantum cryptography using any twononorthogonal states. Phys. Rev. Lett. , 3121 (1992).[5] N. Gisin, G. Ribordy, W. Tittel, and H. Zbinden. Quan-tum cryptography. Rev. Mod. Phys. , 145 (2002).[6] P. Zanardi,Virtual Quantum Subsystems,Phys. Rev. Lett. 87, 077901 (2001). [7] P. Zanardi, D. A. Lidar, and S. Lloyd, QuantumTensor Product Structures are Observable Induced.Phys. Rev. Lett. 92, 060402 (2004).[8] R. Rabelo, M. Ho, D. Cavalcanti, N. Brunner, and V.Scarani, Device-Independent Certification of EntangledMeasurements. Phys. Rev. Lett. 107, 050502 (2011).[9] M. A. Nielsen and I. L. Chuang. Quantum Computationand Quantum Information(Cambridge University Press,Cambridge, U.K., 2004).[10] Yu Cai, Baichu Yu, Pooja Jayachandran, NicolasBrunner, Valerio Scarani, and Jean-Daniel Bancal.Entanglement for any definition of two subsystems.arXiv:2006.07165v1.[11] K. Feng. Unextendible product bases and 1-factorizationof complete graphs. Disc. App. Math.154