The automorphism group of separable states in quantum information theory. University of Illinois at Chicago, Chicago, Illinois ,

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1 The automorphism group of separable states in quantum information theory Shmuel Friedland, 1, a) Chi-Kwong Li, 2, b) Yiu-Tung Poon, 3, c) 4, d) and Nung-Sing Sze 1) Department of Mathematics, Statistics and Computer Science, University of Illinois at Chicago, Chicago, Illinois , USA 2) Department of Mathematics, College of William & Mary, Williamsburg, VA , USA 3) Department of Mathematics, Iowa State University, Ames, IA 50011, USA 4) Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong (Dated: 23 March 2011) We show that the linear group of automorphism of Hermitian matrices which preserves the set of separable states is generated by natural automorphisms: change of an orthonormal basis in each tensor factor, partial transpose in each tensor factor, and interchanging two tensor factors of the same dimension. We apply our results to preservers of the product numerical range. a) Electronic mail: b) Electronic mail: Li is an honorary professor of the University of Hong Kong, the Taiyuan University of Technology, and the Shanghai University. His research was supported in part by a USA NSF grant, a HK RCG grant, and the Key Disciplines of Shanghai Municipality Grant S This research was done while he was visiting the George Washington University during his SSRL leave from the College of William & Mary in the Fall of c) Electronic mail: ytpoon@iastate.edu; Research supported in part by a USA NSF grant. d) Electronic mail: raymond.sze@inet.polyu.edu.hk; Corresponding author. Research supported in part by a HK RCG grant. 1

2 I. INTRODUCTION One of the main concepts in quantum information theory is entanglement. An entangled state involves at least two subsystem or more. We first discuss the two subsystem H m Hn case, a.k.a. bipartite case. Here M n is the space of n n complex matrices and H n M n is the space of n n complex Hermitian matrices. Denote by D n H n the convex set of positive semi-definite matrices of trace one, i.e. density matrices. Also let S m,n D mn H mn H m Hn be the set of bipartite separable states, i.e. S m,n = conv {A B : A D m and B D n }. Clearly, S m,n is a compact convex set. The set of entangled bipartite states is the complement of separable states in D mn, i.e. D mn \ S m,n. Among the best known applications of entanglement are superdense coding, quantum teleportation and more recently measurement based quantum computation (for review, see e.g. Refs. 6 and 12). This recognition sparked an enormous stream of work in an effort to quantify entanglement in both bipartite and multi-partite settings. Among the different measures of entanglement, the relative entropy of entanglement (REE) is of a particular importance. The REE is defined by (c.f. Ref 14): E R (ρ) = min σ S S(ρ σ ) = S(ρ σ), (1) where S is a the set of multi-partite separable states. E R (ρ) is a convex function on S and is strictly convex on strictly positive definite separable states 4. Hence, the computation of E R (ρ), which is given as the minimum of a convex function, should be in principle easy to compute, i.e. polynomial time algorithm. However, E R is hard to compute in general, since the general characterization of separable states is NP-hard 5. A crucial observation of Peres 11 is that S is invariant under the partial transpose. For example, on H mn H m Hn the partial transpose linear map on the second component PT 2 : H mn H mn is induced by PT 2 (A B) = A B, where B is the transposed matrix of B H n. Hence, if a density matrix C D mn represents a separable state then PT 2 (C) is positive semi-definite. (This condition implies that PT 1 (C) = PT 2 (C) is also positive semi-definite, since the transpose map C C, preserves the trace and the positivity.) It 2

3 was shown in Ref. 7 that for m + n 5, C S m,n if and only if C and PT 2 (C) are density matrices. Unfortunately, the positivity of the partial transpose does not imply separability for m + n 6 (c.f. Ref. 7). Denote by G(n 1,..., n k ) the group of linear automorphisms of Hermitian matrices H N k i=1 H n i which leaves invariant the set of separable states S. The structure of G(m, n) was determined recently in Ref. 1. In this paper we extended the above results to G(n 1,..., n k ) for k 3. We show that this group is generated by unitary change of basis in each component, partial transposes in each component, and by permutations of the factors of the same dimension. In summary, G(n 1,..., n k ) consists only of the natural elements. There are related works 8,9 which study the linear maps on k i=1c n i that preserve the product states, i.e. indecomposable tensors. In these papers, the authors show some structural results similar to our results on the group G(n 1,..., n k ). We now briefly summarize the contents of the paper. In Section 2, we give another proof for the structure theorem of G(m, n) obtained in Ref. 1, and the proof is further extended to determine the structure of G(n 1,..., n k ) in Section 3. In Section 4, we apply our results to preservers of the product numerical range. II. THE BIPARTITE CASE In what follows we use the basic notion of the dimension of a convex set C as a subset of R N, denoted by dim C. It is the minimum of the dimension of an affine space, i.e. a translation of a subspace of R N, which contains C. For a set S R n, denote by conv S the convex set spanned by S. For k-linear spaces U 1,..., U k over a given field F, we denote by k i=1 U i the tensor vector space of dimension k i=1 dim U i. Suppose S i is a proper subset of U i for i = 1,..., k. Then k i=1s i = { k i=1u i : u i S i, i = 1,..., k }. Denote by I m H m the identity matrix. Let H + m and H (1) m denote the set of positive semidefinite matrices and Hermitian matrices of trace one, respectively. So H (1) m 3 is a hyperplane

4 in H m with dim H (1) m = m 2 1 and D m = H + m H (1) m. Denote by P m D m the compact set of all Hermitian rank one matrices of trace one, i.e., the set of pure states. Then P m P n is the set of separable pure states in D mn. Observe that K(S m,n ) = conv (H + m H + n ) H + mn is the cone of positive semi-definite matrices generated by separable states. The following result is well known and we bring its proof for completeness. Lemma 1 The set of separable states S m,n is a convex set, whose extreme points is P m P n. Furthermore, dim S m,n = (mn) 2 1 and 1 mn I mn is an interior point of S m,n. Proof. Clearly, since the set of the extreme points of D m is P m, it follows that S m,n = conv (P m P n ). As P m P n P mn, it follows that P m P n is the set of the extreme points of S m,n. Recall next that 1 m I m is an interior point of D m. Hence 1 mn I mn = ( 1 m I m) ( 1 n I n) is an interior point of S m,n. Since S m,n H (1) mn, it follows that dim S m,n = (mn) 2 1. Lemma 2 Let Φ : D mn D mn be an affine map such that Φ(S m,n ) = S m,n. Then Φ can be extended uniquely to an invertible linear map Ψ : H mn H mn. Proof. First extend Φ to an affine homogeneous map, (of degree one), Ψ : K(S m,n ) K(S m,n ) by letting Ψ(tC) = tψ(c) for any t 0 and C S m,n. Clearly Ψ is affine and homogeneous. Also Ψ(K(S m,n )) = K(S m,n ). Since K(S m,n ) K(S m,n ) = H mn, it follows that Ψ extends to a linear map of H mn to itself. Since I mn is an interior point of K(S m,n ), it follows that dim K(S m,n ) = (mn) 2. Hence, dim Ψ(K(S m,n )) = (mn) 2. We claim that the linear map Ψ is invertible. Otherwise dim Ψ(H mn ) (mn) 2 1, which contradicts the fact that dim Ψ(K(S m,n )) = (mn) 2. The proof of Lemma 2 implies that in order to characterize affine automorphisms of separable bipartite states it is enough to consider linear automorphisms of H m which preserve S m,n. The main result of this section is. Theorem 3 Let Ψ : H mn H mn be a linear map. The following are equivalent. (a) Ψ(P m P n ) = P m P n. 4

5 (b) Ψ(S m,n ) = S m,n. (c) There are unitary U M m and V M n such that (c.1) Ψ(A B) = ψ 1 (A) ψ 2 (B) for A B H m Hn, or (c.2) m = n and Ψ(A B) = ψ 2 (B) ψ 1 (A) for A B H m Hn, where ψ 1 has the form A UAU or A UA U, and ψ 2 has the form B V BV or B V B V. To prove Theorem 3, we need the following lemma which can be viewed as the characterization of linear preservers of pure states. Lemma 4 Suppose ψ : H m H n is linear and satisfies ψ(p m ) P n. Then one of the following holds: (i) there is R P n such that ψ has the form A (Tr A)R. (ii) m n and there is a U M m n with UU = I m such that ψ has the form A U AU or A U A U. Proof. Define a map φ : H m+n H m+n given by φ(b) = φ B 1 B 2 = ψ(b 1) 0 for all B = B 1 B 2 H m+n with B 1 H m. B2 B m B2 B 3 Then φ is linear. In particular, φ(a 0 n ) = ψ(a) 0 m for all A H m. Then ψ(p m ) P n implies rank (φ(a)) 1 whenever rank (A) = 1. If dim φ(h m+n ) = 1, then there exist a rank one Q and a linear functional f on H m+n such that φ(b) = f(b)q. Therefore, Q = R 0 m for some R P n and ψ(a) = g(a)r for all A H m where g(a) = f(a 0 n ). Since ψ(p m ) P n, g(p ) = 1 for all P P m. For A H m, let A = m i=1 λ ip i be the spectral decomposition of A. Then g(a) = m i=1 λ if(p i ) = m i=1 λ i = Tr A. 5

6 If dim ψ(h m ) > 1, by Corollary 2 in Ref. 2, there exist α {1, 1} and S M n such that φ has the form B αs BS or B αs B S. Since φ(a 0 n ) = ψ(a) 0 m, ψ has the form A αu AU or A αu A U, where U the leading m n submatrix of S, i.e., S = U. Since ψ(p m ) P n, if ψ has the form ψ(a) = αu AU, then x (αuu )x = Tr(αU (xx )U) = Tr(ψ(xx )) = 1 for all unit vector x C m. This gives αuu = I m. Hence, n m, α = 1 and UU = I m and the result follows. Proof for the case when ψ(a) = U A U is similar. Proof of Theorem 3. The equivalence of conditions (a) and (b) follows from the fact that P m P n is the set of the extreme points of S m,n and that Ψ is linear. The implication (c) (a) is clear. Suppose (a) holds. We will set Ψ(A B) = φ 1 (A, B) φ 2 (A, B), and show that (φ 1 (A, B), φ 2 (A, B)) = (ψ 1 (A), ψ 2 (B)) for all A and B, or m = n and (φ 1 (A, B), φ 2 (A, B)) = (ψ 2 (B), ψ 1 (A)) for all A and B, where ψ 1 and ψ 2 have some standard form. Below are the technical arguments. First, Lemma 2 yields that Ψ is bijective. Without loss of generality, we assume that m n > 1. Consider the partial traces Tr 1 : H mn H n and Tr 2 : H mn H m on H mn H m Hn defined by Tr 1 (A B) = (Tr A) B and Tr 2 (A B) = (Tr B) A. Clearly Tr 1 and Tr 2 are linear maps. Define two maps φ 1 : (H m, H n ) H m and φ 2 : (H m, H n ) H n by φ 1 (A, B) = Tr 2 (Ψ(A B)) and φ 2 (A, B) = Tr 1 (Ψ(A B)). Notice that Ψ(P Q) = φ 1 (P, Q) φ 2 (P, Q) for all P P m and Q P n. (2) Fixed a Q P n, then the maps φ 1 (, Q) : H m H m and φ 2 (, Q) : H m H n are both linear and φ 1 (P m, Q) P m while φ 2 (P m, Q) P n. Therefore, by Lemma 4, both φ 1 (, Q) 6

7 and φ 2 (, Q) have one of the following forms: (i.a) A U AU, (i.b) A U A U, or (ii) A (Tr A) R, (3) where the unitary U and projection R depend on Q. Furthermore, the map φ 2 (, Q) can only be of the form (ii) if m > n. For 1 i, j m, let E ij M m have 1 at the (i, j) entry and 0 elsewhere. Let A = E 11 E 22. Define F : P n R by F (Q) = φ 1 (A, Q), where is the Frobenius norm. Notice that 2 if φ1 (, Q) has the form (i.a) or (i.b), F (Q) = φ 1 (A, Q) = 0 if φ 1 (, Q) has the form (ii). Now for two distinct Q 1, Q 2 P n, write Q 1 = xx and Q 2 = yy with unit vectors x, y C n. Note that x and y are linearly independent. For any t [0, 1], define Q(t) = 1 x + t(y x) 2 (x + t(y x)) (x + t(y x)) P n. In particular, Q(0) = Q 1 and Q(1) = Q 2. For each t [0, 1], as φ 1 (, Q(t)) has the form (i) (i.e. either (i.a) or (i.b)) or (ii), the continuous map t F (Q(t)) is constant. Therefore, one can conclude that either φ 1 (, Q) has the form (i) for all Q P m or φ 1 (, Q) has the form (ii) for all Q P m. Now we claim that one of the following holds. (I) For all Q P n, φ 1 (, Q) has the form (i) and φ 2 (, Q) has the form (ii). (II) For all Q P n, φ 1 (, Q) has the form (ii) and φ 2 (, Q) has the form (i). Suppose first that for some Q P n, both φ 1 (, Q) and φ 2 (, Q) are of the form (i). Then we must have m = n. Then for r = 1, 2, there is unitary matrix U r such that φ r (, Q) has the form A Ur AU r or A Ur A U r. Since m = n 2, the right-hand side of (2) is a quadratic function in P P m while the left-hand side is linear in P P m, which is impossible. To be more precise, let P 1 = E 11, P 2 = E 22, P 3 = 1 2 (E 11+E 12 +E 21 +E 22 ), and P 4 = 1 2 (E 11 E 12 E 21 +E 22 ). (4) 7

8 Then Ψ(P j Q) = U (P j P j )U for all 1 j 4, where U = U 1 U 2. Notice that P 1 + P 2 = P 3 + P 4 and hence P 1 Q + P 2 Q = P 3 Q + P 4 Q. But then Ψ(P 1 Q+P 2 Q) = U (P 1 P 1 +P 2 P 2 )U U (P 3 P 3 +P 4 P 4 )U = Ψ(P 3 Q+P 4 Q), which is a contradiction. Now suppose that for some Q P n, both φ 1 (, Q) and φ 2 (, Q) are of the form (ii). Then φ 1 (A, Q) = (Tr A) R 1 and φ 2 (A, Q) = (Tr A) R 2 for some R 1 P m and R 2 P n. Therefore, Ψ(P Q) = R 1 R 2 for all P P m. This contradicts the fact that Ψ is a bijective map. Therefore, either (I) or (II) holds. Applying a similar argument on the map φ 2 (P, ), one can show that (III) For all P P m, φ 1 (P, ) has the form (ii) and φ 2 (P, ) has the form (i). (IV) For all P P m, φ 1 (P, ) has the form (i) and φ 2 (P, ) has the form (ii). Fix P 0 P m and Q 0 P n. Suppose (I) and (IV) hold. Then there exists for any P P m and Q P n, φ 2 (P, Q) = φ 2 (P 0, Q) = φ 2 (P 0, Q 0 ). Notice that the former equality is by (I) while the latter equality is by (IV). Contradiction arrived. Similarly, it is impossible that both (II) and (III) hold. Hence, we can conclude that either (I) and (III) hold or (II) and (IV) hold. Now suppose (I) and (III) hold. Then ψ 1 ( ) = φ 1 (, Q 0 ) and ψ 2 ( ) = φ 2 (P 0, ) are both of the form (i.a) or (i.b). For all P P m and Q P n, φ 1 (P, ) and φ 2 (, Q) are both of the form (ii). Hence, φ 1 (P, Q 0 ) = φ 1 (P, Q) and φ 2 (P, Q) = φ 2 (P 0, Q). Therefore, Ψ(P Q) = φ 1 (P, Q) φ 2 (P, Q) = φ 1 (P, Q 0 ) φ 2 (P 0, Q) = ψ 1 (P ) ψ 2 (Q). Then by linearity of Ψ and the fact that P m P n spans H mn, the result follows. Finally, if (II) and (IV) hold, we may replace Ψ by the linear map A B Ψ(B A) and apply the above argument. 8

9 III. EXTENSION TO MULTI-PARTITE SYSTEMS One can extend Theorem 3 to tensor product of more than two factors as follows: Theorem 5 Suppose n 1 n k 2 are positive integers with k > 1 and N = k i=1 n i. Assume that Ψ : H N H N ( k i=1 H n i ) is a linear map. The following are equivalent. (a) Ψ ( k i=1p ni ) = k i=1 P ni. (b) Ψ ( conv ( k i=1p ni ) ) = conv ( k i=1p ni ). (c) There is a permutation π on {1,..., k} and linear maps ψ i on H ni for i = 1,... k such that Ψ ( k i=1a i ) = k i=1 ψ i ( Aπ(i) ) for k i=1 A k k i=1p ni, where ψ i has the form X U i XU i or X U i X U i, for some unitary U i M ni and n π(i) = n i for i = 1,..., k. Proof. The implications (c) (a) (b) are clear. Assume that (a) holds. A straightforward generalization of Lemma 2 yields that Ψ is bijective. For 1 r 1 < < r p k, define the following linear map Tr r 1,...,r p : k i=1 H ni p j=1 H nrj k i=1 A i Tr A i p j=1 A r j. i r 1,...,r p In particular, the linear map Tr r : H N H nr is given by Tr ( ) ( ) r k i=1a i = i r Tr(A i) A r. For r = 1,..., k, define maps φ r : (H n1,..., H nk ) H nr by φ r (A 1,..., A k ) = Tr r ( Ψ ( k i=1a i )) for all (A 1,..., A k ) (H n1,..., H nk ). Notice that Ψ ( k i=1p i ) = k r=1 φ r (P 1,..., P k ) for all (P 1,..., P k ) (P n1,..., P nk ). Given arbitrary Q i P ni for i = 2,..., k, the map φ r (, Q 2,..., Q k ) maps P n1 into P nr. By Lemma 4, the map must have the form (i) or (ii) in (3). We claim the following. 9

10 Claim All but one of the maps φ r (, Q 2,..., Q k ), r = 1,..., k, have the form (ii) for all Q i P ni and the exceptional map has and the form (i) for all Q i P ni. Let A 1 = E 11 E 22 H n1. Define F r : (P n2,..., P nk ) R by F r (Q 2,..., Q k ) = φ r (A 1, Q 2,..., Q k ). Similar to the argument in the proof of Theorem 3, F r is a constant function. Thus, either φ r (, Q 2,..., Q k ) always have the form (i) for all Q i P ni, or φ r (, Q 2,..., Q k ) always have the form (ii) for all Q i P ni. Next, since Ψ is a bijection, it is impossible to have all φ r (, Q 2,..., Q k ) being constant maps. Assume that the maps φ s (, Q 2,..., Q k ) and φ t (, Q 2,..., Q k ), with s t, have the form (i) and the rest have the form (ii). In this case, n s = n t = n 1. Consider the linear map L : H n1 H ns Hnt defined by L(A) = Tr s,t ( Ψ ( A ( k i=2q i ))). Then L(P ) = φ s (P, Q 2,..., Q k ) φ t (P, Q 2,..., Q k ) for all P P n1. Recall that φ s (P, Q 2,..., Q k ) and φ t (P, Q 2,..., Q k ) are of the form (i). Following the same argument as in the proof of Theorem 3, one sees that P 1 +P 2 = P 3 +P 4 while L(P 1 )+L(P 2 ) L(P 3 )+L(P 4 ), where P 1, P 2, P 3, and P 4 are defined in (4). This contradicts that L is a linear map. Thus, the claim holds. For p = 2,..., k, applying the same argument on the map φ r (Q 1,..., Q p 1,, Q p+1,..., Q k ), one can show that all but one of the map φ r (Q 1,..., Q p 1,, Q p+1,..., Q k ) have the form (ii) for all Q i P ni and the exceptional map has and the form (i) for all Q i P ni. Furthermore, there is a permutation (π(1),..., π(k)) of (1,..., k) such that φ π(p) (Q 1,..., Q p 1,, Q p+1,..., Q k ) has the form (i) for all Q i P ni. Otherwise, there is r such that φ r (Q 1,..., Q p 1,, Q p+1,..., Q k ) has the form (ii) for all p and for all Q i P ni, which contradicts that Ψ is a bijection. Notice also that n p n π(p) for all p = 1,..., k. This is possible only when n p = n π(p) for all p. Now replacing Ψ by the map of the form k i=1q i Ψ ( k i=1q π (i)) 1, we may assume 10

11 that π(p) = p. Then φ p (Q 1,..., Q p 1,, Q p+1,..., Q k ) has the form (i) for all Q i P ni, and for any r p, φ r (Q 1,..., Q p 1,, Q p+1,..., Q k ) has the form (ii) for all Q i P ni. Now fix some Q i P ni. Then for any P i P ni, Ψ ( k i=1p i ) = k i=1 φ i (P 1,..., P k ) = k i=1φ i (Q 1,..., Q i 1, P i, Q i+1,..., Q k ) = k i=1φ i (P i ), where φ i ( ) = φ i (Q 1,..., Q i 1,, Q i+1,..., Q k ) has the form (i). By the linearity of Ψ, the result follows. Next, we show that one cannot replace condition (b) in Theorem 5 by the weaker condition that Ψ preserves the separable states S = conv ( k i=1p ni ), i.e., Ψ(S) S. In fact, we will see that the convex set L of separable states preserving linear maps has dimension N 4 N 2, which is the dimension of the convex set of density matrices preserving linear maps on H N. Lemma 6 Let H N k i=1 H n i. Define the linear map L 0 : H N H N by L 0 (A) = 1 N Tr(A)I N and let L 1 : H N H N be any linear operator satisfying Tr(L 1 (A)) = 0 for all A H N. Then there exists τ = τ(l 1 ) > 0 such that (L 0 + tl 1 )(S) S for each t ( τ(l 1 ), τ(l 1 )). Furthermore det(l 0 + tl 1 ) = t N 2 1 f(l 1 ), where f(l 1 ) is a minor of order N 2 1 of the representation matrix of L 1 in a basis of H N which contains I N. In particular, if f(l 1 ) 0 then for each t ( τ(l 1 ), τ(l 1 )) \ {0} the linear operator L 0 + tl 1 is invertible. Proof. Clearly, for each t R the operator L(t) = L 0 + tl 1 is trace preserving. Hence it maps the hyperplane Tr(A) = 1 to itself. Note that L(0)(S) = 1 N I N. The generalized version of Lemma 1 yields that dim S = N 2 1 and 1 N I N is an interior point of S. The continuity argument yields that there exists τ = τ(l 1 ) such that (L 0 + tl 1 )(S) lies in the interior of S for t < τ(l 1 ). Let L 1 be the adjoint operator of L 1 with respect to the standard inner product A, B = Tr AB on H N. The assumption that Tr(L 1 (A)) = 0 for all A is equivalent to the assumption 11

12 that L 1 (I N ) = 0. Note that L 0 = L 0, rank L 0 = 1 and L 0 (I N ) = I N. Choose a basis in H N where I N is one of the elements of this basis. Then L 0 = E ii, for some i {1,..., N 2 } and L 1 has a zero row i. Clearly det L(t) = tf(l 1 ) where f(l 1 ) is corresponding minor of L 1. The last claim of the lemma is obvious. Corollary 7 Let L be the set of all linear transformations L : H N H N satisfying L(S) S. Then L is a convex compact set of dimension N 4 N 2. Furthermore the subset L 0 L of invertible transformations is an open dense set in L. Hence dim L 0 = N 4 N 2. Proof. Since any L L is trace preserving it follows that L (I N ) = I N. Let L 1 be the affine set of all linear transformations of H N to itself satisfying L (I N ) = I N. Then L 1 is a translation of a linear subspace of dimension N 4 N 2. Hence dim L N 4 N 2. Lemma 6 yields that dim L = dim L 0 = N 4 N 2. IV. THE PRODUCT NUMERICAL RANGE In Ref. 3 the authors introduced the concept of (tensor) product numerical range of T M mn defined by W (T ) = {Tr(T X) : X P m P n }. This is also known as the decomposable numerical range associated with the tensor product of an operator; see Ref. 10 and its references. It was shown in Ref. 3 that the product numerical range is a useful concept in studying various problems in quantum information theory. To avoid the nontrivial case we let m, n 2. Observe that H m is real subspace of M m and M m = H m 1H m. Hence, any real linear automorphism of H m lifts to a complex linear automorphism of M m. Recall that M m is endowed with the standard inner product X, Y = Tr XY. Assume that Φ : M m M m is a linear map. Then Ψ : M m M m is the dual linear map given by the equality Ψ(X), Y = X, Ψ(Y ) for all X, Y M m. Theorem 3 yields. 12

13 Theorem 8 Let m, n 2 and Ψ : M mn M mn be a linear map. The following are equivalent. (a) W (Ψ (T )) = W (T ) for all T M mn. (b) conv {W (Ψ (T ))} = conv {W (T )} for all T M mn. (c) Ψ has the form described in Theorem 3 (c). Proof. The implications (c) (a) (b) are clear. Suppose (b) holds. Note that conv {W (T )} = {Tr(T Z) : Z S m,n }. Thus the dual map Ψ satisfies Ψ (S m,n ) = S m,n and has the form described in Theorem 3 (c). One readily checks that the dual map of such a map has the same form. The result follows. In the multi-partite case, we can define the product numerical range of a matrix by and deduce the following from Theorem 5. W (T ) = { Tr(T Z) : Z k i=1p ni }, Theorem 9 Suppose n 1 n k 2 are positive integers with k > 1 and N = k i=1 n i > 1. Suppose Ψ : M N M N is a linear map. The following are equivalent. (a) W (Ψ (T )) = W (T ) for all T M N. (b) conv {W (Ψ (T ))} = conv {W (T )} for all T M N. (c) Ψ has the form described in Theorem 5 (c). REFERENCES 1 E. Alfsen and F. Shultz, Unique decompositions, faces, and automorphisms of separable states, Journal of Mathematical Physics 51 (2010), E.M. Baruch and R. Loewy, Linear preservers on spaces of hermitian or real symmetric matrices, Linear Algebra Appl. 183 (1993),

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