PHYSICAL REVIEW A 73, 052314 共2006兲 Quantum separability and entanglement detection via entanglement-witness search and global optimization Lawrence M. Ioannou and Benjamin C. Travaglione Centre for Quantum Computation, Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Wilberforce Road, Cambridge CB3 0WA, United Kingdom 共Received 27 February 2006; published 23 May 2006兲 We focus on determining the separability of an unknown bipartite quantum state by invoking a sufficiently large subset of all possible entanglement witnesses given the expected value of each element of a set of mutually orthogonal observables. We review the concept of an entanglement witness from the geometrical point of view and use this geometry to show that the set of separable states is not a polytope and to characterize the class of entanglement witnesses 共observables兲 that detect entangled states on opposite sides of the set of separable states. All this serves to motivate a classical algorithm which, given the expected values of a subset of an orthogonal basis of observables of an otherwise unknown quantum state, searches for an entanglement witness in the span of the subset of observables. The idea of such an algorithm, which is an efficient reduction of the quantum separability problem to a global optimization problem, was introduced by 关Ioannou et al., Phys. Rev. A 70, 060303共R兲兴, where it was shown to be an improvement on the naive approach for the quantum separability problem 共exhaustive search for a decomposition of the given state into a convex combination of separable states兲. The last section of the paper discusses in more generality such algorithms, which, in our case, assume a subroutine that computes the global maximum of a real function of several variables. Despite this, we anticipate that such algorithms will perform sufficiently well on small instances that they will render a feasible test for separability in some cases of interest 共e.g., in 3 ⫻ 3 dimensional systems兲. DOI: 10.1103/PhysRevA.73.052314 PACS number共s兲: 03.67.Mn, 03.65.Ud II. GEOMETRY OF VECTOR SPACE OF HERMITIAN OPERATORS I. INTRODUCTION Deciding whether a quantum state, be it physical or theoretical, is separable 共as opposed to entangled兲 is a problem of fundamental importance in the field of quantum information processing and is a computationally intractable problem 关1兴. One way to decide that a state is entangled is to use an entanglement witness 共EW兲 关2,3兴. Much work has been done on entanglement witnesses and their utility in investigating the separability of quantum states, e.g., 关4,5兴. EWs have been found to be particularly useful for experimentally detecting the entanglement of states of the particular form p兩典具兩 + 共1 − p兲, where 兩典 is an entangled state and is a mixed state close to the maximally mixed state and 0 艋 p 艋 1 关6,7兴. We will show that the set of separable states is not a polytope and thus there is no finite set of EWs that can detect every entangled state. This work focuses on the principle of invoking a sufficiently large subset of all possible EWs given the expected values of a set of observables. Section II summarizes some geometric aspects of the set of Hermitian operators and Sec. III reviews the geometry of separable states and entanglement witnesses. The simplest case of a set of expected values giving rise to more than one EW is characterized in Sec. IV and illustrated by considering the problem of deciding whether a noisy Bell state is entangled. In Sec. V, we apply the above principle to the problem of detecting the entanglement of a completely unknown quantum state and, in Sec. VI, we outline a class of 共classical兲 algorithms that search for an EW that detects the state or conclude that so such EW exists 共given the currently available information about the state兲. 1050-2947/2006/73共5兲/052314共7兲 Let HM,N denote the set of all Hermitian operators mapping C M 丢 CN to itself. This vector space is endowed with the Hilbert-Schmidt inner product 具X , Y典 ⬅ tr共AB兲, which induces the corresponding norm 储X储 ⬅ 冑tr共X2兲 and distance measure 储X − Y储. By fixing an orthogonal Hermitian basis for HM,N, the elements of HM,N are in one-to-one correspon2 2 dence with the elements of the real Euclidean space R M N . Let B = 兵Xi : i = 0 , 1 , . . . , M 2N2 − 1其 be an orthonormal, Hermit1 ian basis for H M,N, where X0 ⬅ 冑MN I. For concreteness, we can assume that the elements of B are tensor products of the 共suitably normalized兲 canonical generators of SU共M兲 and SU共N兲, given, e.g., in 关8兴. Note that tr共Xi兲 = 0 for all i ⬎ 0. 2 2 Define v : H M,N → R M N −1 as 冤 冥 tr共X1A兲 v共A兲 ª tr共X2A兲 ] . 共1兲 tr共XM 2N2−1A兲 Via the mapping v, the set of separable states SM,N can be 2 2 viewed as a full-dimensional convex subset of R M N −1, 兵v共兲 苸 R M 2N2−1 : 苸 SM,N其, 共2兲 2 2 which properly contains the origin v共I M,N兲 = 0̄ 苸 R M N −1 共recall that there is a ball of separable states of nonzero radius centered at the maximally mixed state I M,N 关9兴兲. 052314-1 ©2006 The American Physical Society PHYSICAL REVIEW A 73, 052314 共2006兲 L. M. IOANNOU AND B. C. TRAVAGLIONE Most of the definitions in the rest of this section may be found in 关10兴. If A 苸 HM,N and A ⫽ 0 and a 苸 R, then 兵x 苸 HM,N : tr共Ax兲 艋 a其 is called the half-space HA,a. The boundary 兵x 苸 HM,N : tr共Ax兲 = a其 of HA,a is the hyperplane A,a with normal A. Call two hyperplanes parallel if they share the ⴰ denote the interior HA,a \ A,a of HA,a. same normal. Let HA,a ⴰ Note that H−A,−a is just the complement of HA,a. For example, the density operators of an M ⫻ N quantum system lie on the hyperplane I,1, where I is the identity operator. Let D M,N = 兵 苸 HM,N : 艌 0其 艚 I,1 denote the density operators. The intersection of finitely many half-spaces is called a polyhedron. Every polyhedron is a convex set. Let D be a polyhedron. A set F 債 D is a face of D if there exists a half-space HA,a containing D such that F = D 艚 A,a. If v is a point in D such that the set 兵v其 is a face of D, then v is a vertex of D. A facet of D is a nonempty face of D having dimension one less than the dimension of D. A polyhedron that is contained in a hyperball 兵x 苸 HM,N : tr共x2兲 = R2其 of finite radius R is a polytope. K 艛 HAi,ai , i=1 where ai ª a*共Ai兲. This would imply that SM,N is K 艚 H−Ai,−ai , i=1 that is, that SM,N is the intersection of finitely many half2 2 spaces. Invoking the isomorphism between HM,N and R M N , 2 2 this says that SM,N is a polytope in R M N −1. Minkowski’s theorem 关10兴 says that every polytope in Rn is the convex hull of its finitely many vertices 共extreme points兲. Recall that an extreme point of a convex set is one that cannot be written as a nontrivial convex combination of other elements of the set. To show that SM,N is not a polytope, it suffices to show that it has infinitely many extreme points. The extreme points of SM,N are precisely the product states, as we now remind ourselves 共see also 关11兴兲: A mixed state is not extreme, by definition. Conversely, we have that III. SEPARABLE STATES AND ENTANGLEMENT WITNESSES 兩典具兩 = 兺 pi兩i典具i兩 共4兲 1 = 兺 pi具兩兩i典具i兩兩典 = 兺 pi兩具i兩典兩2 , 共5兲 i The set of bipartite separable quantum states SM,N in HM,N is defined as the convex hull of the separable pure states 兵兩␣典具␣兩 丢 兩典具兩 苸 HM,N其␣,, where 兩␣典 is a unit vector in C M and 兩典 is a unit vector in CN. Let EM,N = D M,N \ SM,N be the set of entangled states. For each entangled state there ⴰ contains but exists a half-space HA,a whose interior HA,a contains no member of SM,N 关2兴. Call A 苸 HM,N an entanglement witness 关3兴 if for some a 苸 R ⴰ SM,N 艚 HA,a =쏗 and ⴰ EM,N 艚 HA,a ⫽ 쏗. 共3兲 The entanglement witnesses A with a = 0 in 共3兲 correspond to the conventional definition of “entanglement witness” found in the literature, e.g., 关6兴. Entanglement witnesses can be used to determine that a physical quantum state is entangled. Suppose A is an EW as in 共3兲 and that a state that is produced in the laboratory is not known to be separable. If sufficiently many copies of may be produced, then repeatedly measuring the observable A of gives a good estimate of the expected value of A, 具A典 ª tr共A兲, ⴰ which, if less than a, indicates that 苸 HA,a and hence that is entangled. Otherwise, if 具A典 艌 a, then may be entangled or separable. The best value of a to use in 共3兲 is a* = min兩典具兩苸SM,N兵具兩A兩典其 since, with this value of a, the hyperplane A,a is tangent to SM,N and thus the volume of entangled states that can be detected by measuring observable A is maximized. With this in mind, define a*共A兲 ª min 兩典具兩苸S M,N 兵具兩A兩典其 if A is an EW. Detection of the entanglement of reproducible physical states in the laboratory would be straightforward if there were a relatively small number K of EWs Ai such that EM,N is contained in implies i i which implies that 円具i 兩 典円 = 1 for all i; thus, a pure state is extreme. Since SM,N has infinitely many pure product states, we have the following fact, which settles a problem mentioned in 关12兴. 2 2 Fact 1. SM,N is not a polytope in R M N −1. IV. AMBIDEXTROUS ENTANGLEMENT WITNESSES Suppose that A is not an entanglement witness but that −A is. In this case, an estimate of tr共A兲 is just as useful in testing whether is entangled. We extend the definition of “entanglement witness” to reflect this fact: Call A 苸 H M,N a left 共entanglement兲 witness if 共3兲 holds for some a 苸 R, and a right 共entanglement兲 witness if ⴰ SM,N 艚 H−A,−b =쏗 and ⴰ EM,N 艚 H−A,−b ⫽쏗 共6兲 for some b 苸 R. As well, for A a right witness, define b*共A兲 ª max 兩典具兩苸S M,N 兵具兩A兩典其. Note that A is a left witness if and only if −A is a right witness. The operator A 苸 H M,N defines the family 兵A,a其a苸R of 2 2 parallel hyperplanes in R M N . Consider the hyperplane A ª A,tr共A兲/MN which cuts through SM,N at the maximally mixed state I MN. When can A be shifted parallel to its normal so that it separates SM,N from some entangled states? If A is both a left and right witness, then A can be shifted in either the positive or negative direction of the normal. In this case, the two parallel hyperplanes A,a*共A兲 and A,b*共A兲 sandwich SM,N with some entangled states outside of the sand- 052314-2 PHYSICAL REVIEW A 73, 052314 共2006兲 QUANTUM SEPARABILITY AND ENTANGLEMENT¼ wich, which we will denote by W共A兲 ª H−A,−a*共A兲 艚 H−A,−b*共A兲. Definition 1. (Ambidextrous entanglement witness). An operator A 苸 H M,N is an ambidextrous 共entanglement兲 witness if it is both a left witness and a right witness. If A is an ambidextrous witness, then is entangled if 具A典 ⬍ a*共A兲 or if 具A典 ⬎ b*共A兲. We can further define a lefthanded witness to be an entanglement witness that is left but not right. Say that two entangled states 1 and 2 are on opposite sides of SM,N if there does not exist a half-space ⴰ contains 1 and 2 but contains no sepaHA,a such that HA,a rable states. Ambidextrous witnesses have the potential advantage over conventional 共left-handed兲 entanglement witnesses that they can detect entangled states on opposite sides of SM,N with the same physical measurement. Entanglement witnesses can be simply characterized by their spectral decomposition. In the following, suppose A MN−1 i兩i典具i兩 with 苸 H M,N has spectral decomposition A = 兺i=0 0 艋 1 艋 . . . 艋 MN−1. Fact 2. The operator A is a left witness if and only if there exists k 苸 关0 , 1 , . . . , MN − 2兴 such that span共兵兩0典 , 兩1典 , . . . , 兩k典其兲 contains no separable pure states and k+1 ⬎ k. Proof. Suppose first that there exists no such k. Then 兩0典 is, without loss of generality, a separable pure state 共because the eigenspace corresponding to 0 must contain a product state兲, so A cannot be a left witness. To prove the converse, suppose that such a k does exist and that k+1 ⬎ k. Define the real function f共兲 ª tr共A兲 on SM,N. Since span共兵兩0典 , 兩1典 , . . . , 兩k典其兲 contains no separable states and k+1 ⬎ k, the function satisfies f共兲 ⬎ 0. Since the set of separable states is compact, there exists a separable state ⬘ that minimizes f共兲. Thus, setting a ª f共⬘兲 gives ⴰ ⴰ SM,N 艚 HA,a = 쏗. As well, EM,N 艚 HA,a ⫽ 쏗 since tr共A兩0典 䊏 ⫻具0兩兲 = 0 ⬍ a, and so A is a left witness. Theorem 1. The operator A is a left or right entanglement witness if and only if 共i兲 there exists k 苸 关0 , 1 , . . . , MN − 2兴 such that span兵兩0典 , 兩1典 , . . . , 兩k典其 contains no separable pure states and k+1 ⬎ k, or 共ii兲 there exists l 苸 关1 , 2 , . . . , MN − 1兴 such that span兵兩l典 , 兩l+1典 , . . . , 兩 MN−1典其 contains no separable pure states and l ⬎ l−1. Theorem 1 immediately gives a method for identifying and constructing entanglement witnesses. Definition 2. (Partial product basis, unextendible product basis). A partial product basis of C M 丢 CN is a set S of mutually orthonormal pure product states spanning a proper subspace of C M 丢 CN. An unextendible product basis of CM 丢 CN is a partial product basis S of C M 丢 CN whose complementary subspace 共spanS兲⬜ contains no product state 关13兴. We can use unextendible product bases to construct ambidextrous witnesses. Suppose B is an unextendible product basis of C M 丢 CN, and let B⬘ be disjoint from B such that B 艛 B⬘ is an orthonormal basis of C M 丢 CN. One possibility is the left witness defined by A⬘ as A⬘ = − 兺 兩典具兩. 共7兲 兩典苸B⬘ As well, we could split B⬘ into BL⬘ and BR⬘ and define an ambidextrous witness A⬙ as A⬙ = − 兺 兩L典具L兩 + 兩L典苸BL⬘ 兺 兩R典具R兩. 共8兲 兩R典苸BR ⬘ Another thing to realize is that spanB may contain an entangled pure state, which can be pulled out and put into a 共+1兲-eigenvalue eigenspace of A⬘. Depending on B 共and the dimensions M , N兲, there may be several mutually orthogonal pure entangled states in spanB whose span contains no product state; let B⬙ be a set of such pure states. Define the ambidextrous witness as A = − 兺 兩典苸B⬘ 兩典具兩 + 兺 兩典具兩. 共9兲 兩典苸B⬙ This suggests the following problem, related to the combinatorial 关14兴 problem of finding unextendible product bases: Given M and N, find all orthonormal bases B for CM 丢 CN such that B is the disjoint union of ⌳L, B̃, ⌳R; span⌳L and span⌳R contain no product state; span共⌳L 艛 ⌳R兲 contains a product state; and min兵兩⌳L兩 , 兩⌳R兩其 is maximal. Such bases may give “optimal” ambidextrous witnesses, which detect the largest volume of entangled states on opposite sides of SM,N. The functions a* and b* are difficult to compute.1 Thus a criticism of constructing witnesses via the spectral decomposition is that even if you can construct the corresponding physical observables, you still have to perform a difficult computation to make them useful. However, most experimental applications of entanglement witnesses are in very low dimensions, where computing a* and b* deterministically is not a problem—it may even be done analytically, as in the example at the end of this section. Ambidextrous entanglement witnesses 共AEWs兲 represent the simplest case of the principle of invoking as many 共left兲 entanglement witnesses as possible given the expected values of each element of a set X of linearly independent observables, that is, the case 兩X兩 = 1. In Sec. V, we see how this principle generalizes to 兩X兩 ⬎ 1. A simple illustration of how AEWs may be used involves detecting and distinguishing noisy Bell states. Define the four Bell states in C2 丢 C2: 兩±典 ª 共兩00典 ± 兩11典兲/冑2, 兩±典 ª 共兩01典 ± 兩10典兲/冑2. It is straightforward to show that the Bell states are, pairwise, on opposite sides of S2,2. Suppose a left entanglement witness W, with a*共W兲 = 0, detects 兩+典 and 兩+典. Without loss of generality, W can be written in the Bell basis 兵兩+典 , 兩+典 , . . . 其 as 1 At least, 共see Sec. VI兲 WOPT共SM,N兲 is an NP-hard problem 关28兴, because WSEP共S M,N兲 is both NP hard 关1,20兴 and, as the existence of the algorithms described in Sec. VI proves, efficiently reducible to WOPT共S M,N兲. 052314-3 PHYSICAL REVIEW A 73, 052314 共2006兲 L. M. IOANNOU AND B. C. TRAVAGLIONE W= 冤 − ⑀1 a + bi ⫻ ⫻ a − bi − ⑀2 ⫻ ⫻ ⫻ ⫻ ⫻ ⫻ ⫻ ⫻ ⫻ ⫻ 冥 tangled because 苸 E2,2 if one of the following four inequalities is true: , 共10兲 具1 丢 1典 ± 具2 丢 2典 ⬍ − 1/2. for ⑀1 and ⑀2 both positive. But the states 兩s 典 ⬅ 冑12 共兩+典 ± 兩+典兲 are separable. Requiring 具s+兩W兩s+典 艌 0 gives 2a 艌 ⑀1 + ⑀2 and requiring 具s−兩W兩s−典 艌 0 gives 2a 艋 −⑀1 − ⑀2, which, together, give a contradiction. Similar arguments hold for the other pairs of Bell states. Define the operators ± A ª − 兩−典具−兩 + 兩+典具+兩, A ª − 兩−典具−兩 + 兩+典具+兩. Both A and A are easily seen to be AEWs. It is also straightforward to compute the values a*共A兲 = a*共A兲 = − 1/2 and b*共A兲 = b*共A兲 = + 1/2. Suppose that there is a source that repeatedly emits the same noisy Bell state and that we want to decide whether is entangled. Define the Pauli operators: 1 0 ª 冑2 共兩0典具0兩 + 兩1典具1兩兲, 1 ª 冑2 共兩0典具1兩 + 兩1典具0兩兲, 1 2 ª − 3 ª 具1 丢 1典 ± 具2 丢 2典 ⬎ 1/2, i 冑2 共兩0典具1兩 − 兩1典具0兩兲, 1 冑2 共兩0典具0兩 − 兩1典具1兩兲, where 兵兩0典,兩1典其 is the standard orthonormal basis for C2. Noting that A = 1 丢 1 − 2 丢 2 , A = 1 丢 1 + 2 丢 2 , measuring the expected value of the two observables 1 丢 1 and 2 丢 2 may be sufficient to decide that is en- 共11兲 If the noise is known to be of a particular form, then we can also determine which noisy Bell state was being produced. Let 兩B典 be a Bell state. Suppose is known to be of the form p兩B典具B兩 + 共1 − p兲 for some inside both sandwiches W共A兲 and W共A兲. With so defined, one of the four inequalities 共11兲 holds only if exactly one of them holds, so that 兩B典 is determined by which inequality is satisfied. We remark that, if and 兩B典 are known, knowledge of the expected value of any single observable A may allow one to compute p and hence an upper bound on the l2 distance between and the maximally mixed state I / 4. This distance may be enough information to conclude that is separable by checking if is inside the largest separable ball centered at I / 4 关9兴. V. DETECTING ENTANGLEMENT OF AN UNKNOWN STATE USING PARTIAL INFORMATION We now consider the task of trying to decide whether a completely unknown physical state , of which many copies are available, is entangled. For simplicity, we restrict to 苸 H2,2 but the discussion can be applied to a bipartite system of any dimension, replacing Pauli operators with canonical generators of SU共M兲 and SU共N兲 共or any linearly independent Hermitian product basis兲. For such , this problem has already been addressed in 关15兴, where the so-called structural physical approximation of an unphysical map 关16兴 was used to implement the Peres-Horodecki positive partial transpose 共PPT兲 test 关2,17兴. While the structural physical approximation is experimentally viable in principle, it is currently very difficult to do so. Thus, the easiest way to test for entanglement at present is to perform “state tomography” in order to get good estimates of 15 real parameters that define , then reconstruct the density matrix for and carry out the PPT test 关17兴 on this matrix. An experimentalist has many choices of which 15 parameters to estimate: the expectations of any 15 linearly independent observables qualify, as do the probability distributions of any five mutually unbiased 共four-outcome兲 measurements 关18,19兴. Whatever 15 parameters are chosen, we assume that the basic tool of the experimentalist is the ability to perform local two-outcome measurements on each qubit, e.g., measuring 1 on the first qubit and 2 on the second. Under this assumption, the scenario where the two qubits of are far apart is easily handled if classical communication is allowed between the two laboratories. We further assume, for simplicity, that the set of these local twooutcome measurements is the set of Pauli operators 兵i其i=0,1,2,3 共defined previously兲. If i is measured on the first qubit and j on the second, repeating this procedure on many copies of gives good estimations of the three expectations 具i 丢 0典, 具0 丢 j典, and 具i 丢 j典 共where the subscript is omitted for readability兲. Let us call this procedure measuring i j . 052314-4 PHYSICAL REVIEW A 73, 052314 共2006兲 QUANTUM SEPARABILITY AND ENTANGLEMENT¼ Suppose the experimentalist sets out to solve our problem and begins the data collection by measuring 11 and then 22. Even though only six of the 15 independent parameters defining have been found, the example in the previous section shows that is entangled if one of the four inequalities 共11兲 is true. It is straightforward to show that if none of these inequalities is true, then no entanglement witness in the span of 兵1 丢 1 , 2 丢 2其 can detect if it is entangled.2 However, there may be an entanglement witness in the span of want to apply the entanglement witness optimization procedure 关4兴 to the result of the algorithm, as these algorithms do not necessarily output optimal entanglement witnesses兲. Let j be the number of nontrivial expected values of that are known, 2 艋 j 艋 M 2N2 − 1; that is, 共without loss兲 assume we know the expected values of the elements of B⬘ = 兵X1 , X2 , . . . , X j其. The algorithms either find an entanglement witness in span共B⬘兲 for , or conclude that no such witness M 2N2−1 exists. For any Y 苸 H M,N with Y = 兺i=0 y iXi, let Ȳ be the j-dimensional vector of the real numbers y i for i = 1 , 2 , . . . , j. Define 兵 0 丢 1, 0 丢 2, 1 丢 1, 2 丢 2, 1 丢 0, 1 丢 0其 SM,N = 兵¯: 苸 SM,N其. that does detect . More generally, at any stage of the data-gathering process, if we have the set of expectations 兵具i 丢 j典 : 共i , j兲 苸 T其, then is entangled if there is an entanglement witness in the span of 兵i 丢 j : 共i , j兲 苸 T其 that detects 关T 傺 {共k , l兲 : k , l 苸 兵0 , 1 , 2 , 3其} \ 共0 , 0兲兴. If the experimentalist has access to a computer program that can quickly discover such an entanglement witness 共if it exists兲, then the data-gathering process can be terminated early and no more qubits have to be used to decide that is entangled. The algorithms described in the next section are just such programs. To see this, note that the projection S2,2 of S2,2 onto span兵i 丢 j : 共i , j兲 苸 T其 is a full-dimensional convex subset of R兩T兩, and the projection ¯ of onto span兵i 丢 j : 共i , j兲 苸 T其 is a point in R兩T兩 such that ¯ 苸 S2,2 if and only if there is an entanglement witness in the span of 兵i 丢 j : 共i , j兲 苸 T其 that detects . Since the following algorithms can be applied to any full-dimensional convex set 共satisfying certain conditions兲, we can apply them to S2,2. We view any such algorithm as an extra tool that an experimentalist can use to facilitate entanglement detection and minimize the number of copies of that must be measured— essentially, trading classical resources for quantum resources. As we saw in the case of constructing ambidextrous witnesses, the primary classical resource required to invoke a sufficiently large subset of all such entanglement witnesses is a subroutine for computing the function b* 共equivalently, a*兲. VI. ALGORITHMS FOR FINDING ENTANGLEMENT WITNESSES BASED ON GLOBAL OPTIMIZATION Assume that 苸 D M,N is a state whose separability is unknown. We can handle two scenarios—one experimental, as described above, and the other theoretical. In the theoretical scenarios, we assume that we know the density matrix for ; this corresponds to having gathered all M 2N2 − 1 independent expected values in the experimental scenario. Since the algorithms find an entanglement witness when 苸 EM,N, they could also be applied when is known to be entangled but an entanglement witness for is desired 共though one may 2 To show this, it suffices to find four separable states whose projections onto span兵1 丢 1 , 2 丢 2其 are the four vertices of the square with vertices 共1 / 2 , 0兲, 共0 , 1 / 2兲, 共−1 / 2 , 0兲, and 共0 , −1 / 2兲; such states are 共1 / 4兲I ± 共1 / 2兲i 丢 i for i = 1 , 2. The result then follows from convexity of S2,2. 共12兲 Note that SM,N is a full-dimensional convex set in R j, properly containing the origin 共since I M,N is the zero vector in R j兲. Let K be a full-dimensional convex subset of Rn which contains a ball of finite nonzero radius centered at the origin and is contained in a ball of finite radius R. The algorithms are general and can be used to decide whether a hyperplane exists which separates a given point p from any given K satisfying these properties. Thus, the clearest way to describe how the algorithms work is to use the application-neutral notation of convex analysis. For x 苸 Rn and ␦ ⬎ 0, let B共x , ␦兲 ª 兵y 苸 Rn : 储x − y储 艋 ␦其. For a convex subset K 傺 Rn, let S共K , ␦兲 ª 艛x苸KB共x , ␦兲 and S共K , −␦兲 ª 兵x : B共x , ␦兲 債 K其. Define the following convex body problems 关20兴. Definition 3. {Weak separation problem for K 关WSEP共K兲兴}. Given a rational vector p 苸 Rn and rational ␦ ⬎ 0, either assert p 苸 S共K , ␦兲, or find a rational vector c 苸 Rn with 储c储⬁ = 1 such that cTx ⬍ cT p for every x 苸 K.3 Definition 4. {Weak optimization problem for K 关WOPT共K兲兴}. Given a rational vector c 苸 Rn and rational ⑀ ⬎ 0, either find a rational vector y 苸 Rn such that y 苸 S共K , ⑀兲 and cTx 艋 cTy + ⑀ for every x 苸 K; or assert that S共K , −⑀兲 is empty.4 By taking ␦ and ⑀ to be zero, we implicitly define the corresponding strong separation 共SSEP兲 and strong optimization 共SOPT兲 problems. Note that by taking K to be SM,N and p to be ¯ 共for some state 苸 D M,N兲, SSEP共K兲 corresponds to the problem of finding an entanglement witness for 共or deciding that one does not exist in the span of B⬘兲;5 and by further taking c to be Ā, for some A 苸 H M,N, SOPT共K兲 corresponds to the problem of computing b*共A兲 关actually something at least as hard, since SOPT共K兲 asks for the maximizer of cTx, over x 苸 K, rather than just the maximum兴. 3 The l⬁ norm appears here as a technicality, so that c need not be normalized by a possibly irrational multiplier. We will just use the Euclidean norm in what follows and have 储c储 ⬇ 1. 4 This will never be the case for us, as S M,N is not empty. 5 If p arises from some estimation procedure 共as in our experimental setting兲, then there is a hyperbox around p that contains the “actual” point p; the hyperbox is given by the error bars on each coordinate of p. From the error bars can be computed a ⌬ ⬎ 0 such that p 苸 B共p , ⌬兲. If the WSEP共K兲 algorithm asserts p 苸 S共K , ␦兲, then we can only assert that p is in S共K , ␦ + ⌬兲; otherwise, we can only assert that cTx ⬍ cT p + ⌬ for every x 苸 K. 052314-5 PHYSICAL REVIEW A 73, 052314 共2006兲 L. M. IOANNOU AND B. C. TRAVAGLIONE We describe oracle-polynomial-time algorithms for WSEP共K兲, with respect to an oracle for WOPT共K兲; that is, assuming each call to the oracle is assigned unit complexity cost, the algorithms have running time in O(poly关n , ln共R / ␦兲兴). We will use “O” to denote oracles 共black-boxed subroutines兲 for problems, indicating which problem via a subscript, e.g., OSOPT共K兲. In what follows, so as not to obfuscate the main idea of the algorithms, we ignore the weakness of the separation and optimization problems; that is, we assume we are solving SSEP共K兲 with an oracle for SOPT共K兲. There are at least two ways to reduce SSEP共K兲 to SOPT共K兲. The first method was covered in 关21,22兴; the second method, which we give below, is well known and may be found in the synthesis of Lemma 4.4.2 and Theorem 4.2.2 in 关20兴. For y 苸 Rn and b 苸 R, define the hyperplane y,b ⬅ 兵x 苸 Rn : y Tx = b其. Definition 5 (Polar of K). The polar K쐓 of a fulldimensional convex set K 傺 Rn that contains the origin is defined as K쐓 ª 兵c 苸 Rn:cTx 艋 1 " x 苸 K其. 共13兲 If c 苸 K쐓, then the plane c,1 separates p 苸 Rn from K when cT p ⬎ 1. Definition 6. {Feasibility problem for K⬘ 关FEAS共K⬘兲兴}. Given a convex set K⬘ 傺 Rn, either find a point k⬘ 苸 K⬘, or assert that K⬘ is empty. Thus, the separation problem for p is equivalent to the feasibility problem for Q p, defined as Q p ª K쐓 艚 兵c:pTc 艌 1其. 共14兲 As outlined in the next section, to solve the feasibility problem for any K⬘, it suffices to have a separation routine for K⬘. Because we can easily build a separation routine OSSEP共Qp兲 for Q p out of OSSEP共K쐓兲, it suffices to have a separation routine OSSEP共K쐓兲 for K쐓 in order to solve the feasibility problem for Q p.6 Building OSSEP共Qp兲 out of OSSEP共K쐓兲 is done as follows: Routine OSSEP共Qp兲共y兲: CASE: pTy ⬍ 1 RETURN −p ELSE: pTy 艌 1 CALL OSSEP共K쐓兲共y兲 CASE: OSSEP共K쐓兲共y兲 returns separating vector q RETURN q ELSE: OSSEP共K쐓兲共y兲 asserts y 苸 K쐓 RETURN “y 苸 Q” It remains to show that the optimization routine OSOPT共K兲 for K gives a separation routine OSSEP共K쐓兲 for K쐓. Suppose y is given to OSOPT共K兲, which returns k 苸 K such that y Tx 艋 y Tk ¬ b for all x 苸 K. If b 艋 1, then OSSEP共K쐓兲 may assert 6 We slightly abuse the oracular O notation by using it for both truly oracular 共black-boxed兲 routines and for other 共possibly not completely black-boxed兲 routines. y 苸 K쐓. Otherwise, OSSEP共K쐓兲 may return k, because k,1 共and hence k,b兲 separates y from K쐓: since kTy = b ⬎ 1, it suffices to note that kTc = cTk 艋 1 for all c 苸 K쐓 by the definition of K쐓 and the fact that k 苸 K. The plane k,1 is called a cutting plane, and, to solve FEAS共K⬘兲 with OSSEP共K⬘兲, we use a cutting-plane algorithm. All such algorithms have the same basic structure. 共1兲 Define a 共possibly very large兲 regular bounded convex set P0 which is guaranteed to contain K⬘, such that, for some reasonable definition of “center,” the centre 0 of P0 is easily computed. The set P0 is called an outer approximation to K⬘. Common choices for P0 are the origin-centered hyperbox 兵x 苸 Rn : −2L 艋 xi 艋 2L, 1 艋 i 艋 n其 and the origin-centered hyperball 兵x : xTx 艋 2L其 共where 2L is a trivially large bound兲. 共2兲 Give the center of the current outer approximation P to OSSEP共K⬘兲. 共3兲 If OSSEP共K⬘兲 asserts “ 苸 K⬘,” then halt. 共4兲 Otherwise, say OSSEP共K⬘兲 returns the cutting plane c,b such that K⬘ 傺 兵x : cTx 艋 b其. Update 共shrink兲 the outer approximation P ª P 艚 兵x : cTx 艋 b⬘其 for some b⬘ 艌 b; the idea is that the new P has about half the volume of the old P 共i.e., usually c,b passes through , or near it兲. Possibly perform other computations to further update P. Check stopping conditions; if they are met, then halt. Otherwise, go to step 2. The difficulty with such algorithms is knowing when to halt in step 4. Generally, the stopping conditions are related to the size of the current outer approximation. Because it is always an approximate 共weak兲 feasibility problem that is solved, the associated accuracy parameter ␦⬘ can be exploited to get a “lower bound” V on the “size” of K⬘, with the understanding that if K⬘ is smaller than this bound, then the algorithm can correctly assert that S共K⬘ , −␦⬘兲 is empty. Thus the algorithm stops in step 4 when the current outer approximation is smaller than V. Using the above reduction from SSEP共K兲 to SOPT共K兲, there are a number of polynomial-time convex feasibility algorithms that can be applied 共see 关23兴 for a discussion of all of them兲. The three most important are the ellipsoid method, the volumetric-center method, and the analyticcenter method. The latter are more efficient than the ellipsoid algorithm and are very similar to each other in complexity and precision requirements, with the analytic-center cuttingplane 共ACCP兲 algorithm in 关23兴 having some supposed practical advantages. We refer to 关22,23兴 共and references therein兲 for a discussion of details, including computer precision requirements, for either of the ACCP algorithms arising from either the reduction in 关21兴 or the well-known one given here. VII. CLOSING REMARKS In terms of attempting to find a practical algorithm for the quantum separability problem, the skeptic notices that such algorithms appear not to offer any advantage over other approaches: instead of having to solve one instance of an NP-hard problem, we now have to solve many. In response to such skepticism, we can, at least, refer to 关22兴, where 052314-6 PHYSICAL REVIEW A 73, 052314 共2006兲 QUANTUM SEPARABILITY AND ENTANGLEMENT¼ optimization 关26兴; and Hansen’s global optimization algorithm using interval analysis 关27兴. it is shown that the asymptotic complexity of such algorithms compares favorably with all other deterministic algorithms 共with known worst-case complexity bounds兲 for the quantum separability problem. 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