RANKS OF QUANTUM STATES WITH PRESCRIBED REDUCED STATES

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1 RANKS OF QUANTUM STATES WITH PRESCRIBED REDUCED STATES CHI-KWONG LI, YIU-TUNG POON, AND XUEFENG WANG Abstract. Let M n be the set of n n complex matrices. in this note, all the possible ranks of a bipartite state in M m M n with prescribed reduced states in the two subsystems, are determined. The results are used to determine the Choi rank of quantum channels Φ : M m M n sending I/m to a specific state σ M n. Key words. Quantum state, Reduced state, Partial trace, Quantum channel, Rank, Eigenvalue. AMS subject classifications. 15A18, 15A60, 15A4, 15B48, 46N Introduction. In quantum information science, quantum states are used to store, process, and transmit information. Mathematically, quantum states are represented by density matrices, i.e., positive semidefinite matrices of trace 1. Let M n (H n ) be the set of n n complex (Hermitian) matrices. Let D n be the set of density matrices in H n. Suppose σ 1 D m and σ D n are two quantum states. Their product state is σ 1 σ D mn. The combined system is known as the bipartite system, and a general quantum state is represented by a density matrix ρ D mn. Two important quantum operations used to extract information of the subsystems from a quantum state of the bipartite system are the partial traces defined as tr 1 (σ 1 σ ) = σ and tr (σ 1 σ ) = σ 1 on tensor states σ 1 σ D mn, and extended by linearity for general states in D mn. In particular, suppose ρ = (ρ ij ) 1 i,j m D mn such that ρ ij M n. Then tr 1 (ρ) = ρ ρ mm M n and tr (ρ) = (trρ ij ) 1 i,j m M m. Let rank (σ) be the rank of a matrix σ. The purpose of this note is to give a complete answer to the following. Problem. Determine all the possible values of rank (ρ) for ρ D mn in the set for given σ 1 D m and σ D n. S(σ 1, σ ) = {ρ D mn : tr 1 (ρ) = σ, tr (ρ) = σ 1 } Received by the editors on November 1, 017. Accepted for publication on May 31, 018. Handling Editor: Panayiotis Psarrakos. Corresponding Author: Chi-Kwong Li. Department of Mathematics, College of William and Mary, Williamsburg, VA 3187, USA (ckli@math.wm.edu). Department of Mathematics, Iowa University, Ames, IA 50011, USA (ytpoon@iastate.edu). School of Mathematical Sciences, Ocean University of China, Qingdao, Shangdong 66100, China (wangxuefeng@ouc.edu.cn).

2 Chi-Kwong Li, Yiu-Tung Poon, and Xuefeng Wang It is well known and easy to verify that the maximum rank of ρ S(σ 1, σ ) equals R = rank (σ 1 )rank (σ ). In Section, we will present a finite algorithm for determining the minimum rank value r of ρ S(σ 1, σ ) in terms of the eigenvalues of σ 1 and σ. Moreover, we will show that there is ρ S(σ 1, σ ) with rank (ρ) = k for every k {r, r + 1,..., R}. In Section 3, we will describe some implications of the results in Section to the study of quantum channels. In particular, the results allow us to determine all the possible Choi ranks of a quantum channel Φ : M m M n having a prescribed state Φ(I m /m) D n. In our discussion, let M m,n be the set of m n complex matrices so that M n = M n,n. Denote by U n and H n the set of unitary matrices and the set of Hermitian matrices in M n, respectively. Let R n denote the set of n-tuples in R n with decreasing coordinates. Given σ H n, λ(σ) = (λ 1 (σ),..., λ n (σ)) R n will denote the eigenvalues of σ arranged in decreasing order. Let {e (m) 1,..., e (m) m } and {e (n) 1,..., e(n) clearly, {e (m) 1 e (n) 1, e(m) 1 e (n),..., e(m) m we use the notation e i for e (m) i e i e j instead of e (m) i e (n) j n } be the standard bases for C m and C n, respectively. Then, e (n) n } is the standard basis for C m C n C mn. For simplicity, or e (n) i if the dimension of the vector is clear in the context. Also, we use, and e i e t j = E ij to denote the basic complex unit of appropriate size. The following observation is useful in our discussion. Lemma 1.1. Let σ 1 D m, σ D n, U M m and V M n. Then S(Uσ 1 U, V σ V ) = (U V )S(σ 1, σ )(U V ) = {(U V )ρ(u V ) : ρ S(σ 1, σ )}.. Ranks of matrices in S(σ 1, σ ). In this section, we present results concerning the ranks of states in S(σ 1, σ ). We will use the fact that S(σ 1, σ ) contains a rank one matrix if and only if σ 1 and σ have the same nonzero eigenvalues counting multiplicities; see [, 5. For w = (w 1,..., w mn ) t C mn, let W = [w be the m n matrix such that the ith row equals (w (i 1)n+1,..., w in ) for i = 1,..., m. Suppose W = [w has singular value decomposition XSY t such that X M m is unitary with columns x 1,..., x m, Y M n is unitary with columns y 1,..., y n and S = s 1 e 1 e t s k e k e t k, where k is the rank of W. It follows that W = [w = k s jx j yj t, and w = k s jx j y j, which is known as the Schmidt decomposition of w. Theorem.1. Let (σ 1, σ ) D m D n such that rank (σ 1 ) = r 1 r = rank (σ ). There is an element in S(σ 1, σ ) attaining the lowest rank r with r r 1. Moreover, there exists ρ S(σ 1, σ ) of rank k if and only if r k r 1 r. Proof. By Lemma 1.1, we may assume that σ 1 = diag(µ 1,..., µ r1, 0,..., 0) and σ = diag(ˆµ 1,..., ˆµ r, 0,..., 0). If ρ = k z jz j S(σ 1, σ ), then tr 1 (ρ) = σ and tr (ρ) = σ 1. For 1 i r 1 and 1 j r, let e i e j denote e (m) i l = 1,..., r 1, let r z l = r1+1 l i=1 v i+l 1,i + r e (n) j, and let v ij = µ i ˆµ j e i e j. Now, for i=1 v i+l 1,i if 1 l r 1 r + 1, i=r 1+ l v i+l 1 r 1,i if r 1 r + 1 < l r 1.

3 3 Ranks of Quantum States With Prescribed Reduced States For example, z 1 = v v r,r, z = v 1 + v v r+1,r,..., z r1 = v r1,1 + v v r 1,r. Then (.1) tr 1 (z l zl ) = r i=1 µ i+l 1 ˆµ i E ii if 1 l r 1 r + 1, r1+1 l i=1 µ i+l 1 ˆµ i E ii + r i=r 1+ l µ i+l 1 r 1 ˆµ i E ii if r 1 r + 1 < l r 1, and (.) tr (z l zl ) = r i=1 µ i+l 1 ˆµ i E i+l 1 i+l 1 if 1 l r 1 r + 1, r1+1 l i=1 µ i+l 1 ˆµ i E i+l 1 i+l 1 + r i=r 1+ l µ i+l 1 r 1 ˆµ i E i+l 1 r1 i+l 1 r 1 if r 1 r + 1 < l r 1. Let (.3) ρ = r 1 l=1 z l z l S(σ 1, σ ). Then ρ has rank r 1. It follows from (.1) and (.) that ρ S(σ 1, σ ). In (.3) if we replace z 1 z 1 by v 11 v v pp vpp + v jj v jj, p = 1,..., r 1, j>p j>p the resulting state is in S(σ 1, σ ) with rank p + r 1 for p = 1,..., r 1. Similarly, we can replace each z j z j by a rank p + 1 matrix for p = 1,..., r 1, in such a way that the resulting state still lies in S(σ 1, σ ). Hence, S(σ 1, σ ) contains matrices of rank k for every k {r 1,..., r 1 r }. Now, suppose ρ S(σ 1, σ ) has rank r < r 1. We will show that there exists ρ S(σ 1, σ ) with rank k for any r < k < r 1. We prove the result by induction on r. To this end, let (.4) ρ = z j zj S(σ 1, σ ). By the Schmidt decomposition, z j = t j l=1 s jlx jl y jl for each j, where t j = rank ([z j ). Let w jl = s jl x jl y jl. Similar to the previous case, we can replace z j zj by w jl wjl + w jl w jl l>p l>p l p so that the resulting state still lies in S(σ 1, σ ). We may increase the number of summands in (.4) by one at each time until we get ˆρ = j,l w jlwjl. Note that in each step, the rank of the state will either stay the same or increase by 1, and rank (ˆρ) rank (tr (ˆρ)) = rank (σ 1 ) = r 1. As a result, the set S(σ 1, σ ) contains matrices of ranks r,..., r 1. We illustrate the above theorem by the following example.

4 Chi-Kwong Li, Yiu-Tung Poon, and Xuefeng Wang 4 Example.. Suppose σ 1 = diag(a 1, a, a 3 ) and σ = diag(b 1, b ) with a 1 a a 3 > 0, b 1 b > 0. Let z 1 = ( a 1 b 1, 0, 0, a b, 0, 0) t, z = (0, 0, a b 1, 0, 0, a 3 b ) t, z 3 = (0, a 1 b, 0, 0, a 3 b 1, 0) t. Then ρ = z 1 z 1 + z z + z 3 z 3 S(σ 1, σ ) is of rank 3. One can replace z 1 z 1 by v 1 v 1 + v v with v 1 = ( a 1 b 1, 0, 0, 0, 0, 0) t and v = (0, 0, 0, a b, 0, 0) t to get a rank 4 matrix in S(σ 1, σ ). Similarly, we can further replace z z by v 3 v 3 + v 4 v 4, and z 3 z 3 by v 5 v 5 + v 6 v 6, etc. to get matrices in S(σ 1, σ ) of rank 5 and 6. Corollary.3. Let (σ 1, σ ) D m D n. There is a rank one ρ S(σ 1, σ ) if and only if σ 1 and σ have the same (multi-)set of non-zero eigenvalues, say, λ 1 λ r. In such a case, there exists ρ S(σ 1, σ ) of rank k if and only if 1 k r. Proof. The first part follows from the fact that ρ = vv S(σ 1, σ ), where v has Schmidt decomposition r s jx j y j, if and only if σ 1 = r s j x jx j and σ = r s j y jyj. The second part follows from Theorem.1. We illustrate the above Corollary by the following example. Example.4. Suppose σ 1 = diag(λ 1, λ, 0) and σ = diag(λ 1, λ ) such that λ 1 λ > 0 and λ 1 + λ = 1. Let f 1 = ( λ 1, 0, 0, 0, 0, 0) t, f = (0, 0, 0, λ, 0, 0) t. Then (f 1 + f )(f 1 + f ) S(σ 1, σ ) has rank 1, and f 1 f 1 + f f S(σ 1, σ ) has rank. Let v 11 = (λ 1, 0, 0, 0, 0, 0) t, v 1 = (0, 0, 0, λ, 0, 0) t, v 1 = (0, 0, λ λ 1, 0, 0, 0) t, v 31 = (0, λ 1 λ, 0, 0, 0, 0) t. Then (v 11 + v 1 )(v 11 + v 1 ) + v 1 v 1 + v 31 v 31 S(σ 1, σ ) has rank 3 and v 11 v 11 + v 1 v 1 + +v 1 v 1 + v 31 v 31 S(σ 1, σ ) has rank 4. Next, we determine the minimal rank of ρ in S(σ 1, σ ). For w = (w 1,..., w mn ) t C mn, we continue to let W = [w be the m n matrix such that the ith row equals (w (i 1)n+1,..., w in ) for i = 1,..., m. One can easily construct the inverse map which converts an m n matrix W to w = vec(w ) C mn so that W = [w. Note that tr 1 (ww ) = W t (W t ) and tr (ww ) = W W. Suppose ρ S(σ 1, σ ) has rank r. Then there exists an mn r matrix V such that tr 1 (V V ) = σ and tr (V V ) = σ 1. For 1 j r, let W j = [v j, where v j is the j th column of V. Then we have σ 1 = W 1 (W 1 ) + +W r (W r ) and σ = W t 1 (W t 1) + +W t r (W t r). Given C H m, let λ(c) = (c 1,..., c m ) denote the eigenvalues of C with c 1 c m. Let R m = {(x i ) R m : x 1 x m }. By a result of Klyachko [4,, the eigenvalues of a sum of Hermitian matrices can be characterized by a set LR(m, r) of r + 1 tuples (J 0, J 1,..., J r ), where J 0,..., J r are subsets of {1,..., m}. More specifically, given a = (a 1,..., a m ) and c j = (c 1j,..., c mj ) R m, 1 j r, there exist A, C 1,..., C r H n such that A = C j with λ(a) = a, and λ(c j ) = c j, 1 j r if and only if tra = r tr (C j) and for every (J 0, J 1,..., J r ) LR(m, r), the inequality

5 5 Ranks of Quantum States With Prescribed Reduced States (.5) holds. We have the following. i J 0 a i c ij i J j Theorem.5. Suppose m n, σ 1 D m and σ D n. The following conditions are equivalent: (1) There exists ρ S(σ 1, σ ) with rank r. () There exist C 1,..., C r H m and C 1,..., C r H n such that (i) λ( C j ) = λ(c j O n m ), (ii) σ 1 = C j, and (iii) σ = C j. (3) There exists C H m such that () holds with λ(c j ) = λ(c) and λ( C j ) = λ(c O n m ) for all 1 j r. Proof. (1) () : Suppose there exists ρ S(σ 1, σ ) with rank r. Then there exist W 1,..., W r M m,n such that σ 1 = W 1 (W 1 ) + + W r (W r ) and σ = W1 t ( ) W t ( ) W t r W t r. Then () holds with C j = W j W j and C j = W t j (W t j ). () (1) can be proven by reversing the above argument. () (3) : Suppose λ(σ 1 ) = (a 1,..., a m ), and λ(c j ) = (c 1j,..., c mj ) for j = 1,..., r. Let (J 0, J 1,..., J r ) LR(m, r), and π = (1,,..., r) be the cyclic permutation of {1,..., r}, i.e., π(j) = j + 1 for 1 j < r and π(r) = 1. Then (J 0, J πk (1),..., J πk (r)) LR(m, r). Since σ 1 = C C r, by (.5) we have i J 0 a i r i J j c iσ k (j) for every 0 k r 1 i J 0 a i r i J j c i, where c i = (c i1 + + c ir )/r σ 1 = Ĉ1 + + Ĉr for some Ĉ1,..., Ĉr H m, where λ(ĉj) = (c 1,..., c m ) for 1 j r. Similarly, we can choose C j such that λ( C j ) = (c 1,..., c m, 0,..., 0). Clearly, (3) (). Remark.6. For every m, n 1, there exists a permutation matrix P M mn such that ρ S(σ 1, σ ) if and only if P ρp T S(σ, σ 1 ). For example, let P = Then for every σ 1 D and σ 1 D 3, ρ S(σ 1, σ ) if and only if P ρp T S(σ, σ 1 ). Thus, there is no loss of generality in the restriction of m n in Theorem.5.

6 Chi-Kwong Li, Yiu-Tung Poon, and Xuefeng Wang 6 Example.7. Suppose m =, n = 3, ρ = 1 1 σ 1 = 1 [ We have and, σ = Let C 1 = 1 [ C 1 = , C = 1 [ , C = (i) λ(c 1 ) = 1 1 (3, 1), λ( C 1 ) = 1 1 (3, 1, 0), λ(c ) = 1 1 (8, 0), λ( C ) = 1 (8, 0, 0). 1 (ii) σ 1 = C 1 + C. (iii) σ = C 1 + C. Let c 1 = 1 ( ) = and c = 1 ( ) = 1. Then we can choose 4 C 1 = 1 [ Ĉ 1 = 1 4 Ĉ = , C = 1 [ , ,,.. Then ρ S(σ 1, σ ) with Then we have σ 1 = C 1+C, σ = Ĉ1+Ĉ with λ(c 1) = λ(c ) = (c 1, c ) and λ(ĉ1) = λ(ĉ) = (c 1, c, 0). When n = rm, we have the following corollary of Theorem.5. Corollary.8. Suppose n = rm, σ 1 D m and σ D n. The following conditions are equivalent: (a) There exists ρ S(σ 1, σ ) with rank r. (b) There exists C = (C ij ) H n with C ij M m such that λ(c) = λ(σ ) and λ (C C rr ) = λ(σ 1 ). (c) Condition (b) holds with λ (C 11 ) = = λ (C rr ).

7 7 Ranks of Quantum States With Prescribed Reduced States Proof. (a) (b) : Suppose (a) holds. Let C j and C j, 1 j r, be as given by condition () in Theorem.5. For each 1 j r, there exists an n n unitary matrix U j such that C ( ) j = U j Cj O (r 1)m U j. Let V j M n,m be formed by the first m columns of U j. Let R j = V j C 1/ j and R = [R 1 R r. Set C = R R. Then and λ(c) = λ(r R) = λ(rr ) = λ λ (C C rr ) = λ Rj R j = λ Therefore, C = R R will satisfy condition (b). C j = λ(σ ) C j = λ(σ 1 ). (b) (a) : Suppose (b) holds. Let C = R R where R = [R 1 R r, with R j M n,m. Then C ii = R i R i and λ(c) = λ(rr ). Hence, we have λ(σ 1 ) = λ (R 1R R rr r ) and λ(σ ) = λ (R 1R R rr r ). Since λ(r i R i) = λ ( R i R i O (r 1)m), the result follows from condition () in Theorem.5. (b) (c) follows from () (3) in Theorem.5. Applying Theorem.5, we have the following algorithm for finding the minimal rank r of matrices in S(σ 1, σ ). Algorithm.9. Suppose σ 1 D m and σ D n, with n m. Let λ(σ 1 ) = (a 1,..., a m ) and λ(σ ) = (b 1,..., b n ). Step 1. If b i = a i for 1 i m, then r = 1. If not, r > 1, then go to step. Step. For r > 1, suppose the previous steps shows that the minimal rank is r. Let m P r (a 1,..., a m ) = c Rm : c 1,..., c m 0, c j = 1/r, a i c i for all (J 0, J 1,..., J r ) LR(m, r), i J 0 i J j m Q r (b 1,..., b n ) = c Rm : c 1,..., c m 0, c j = 1/r, i J 0 b i ĉ i for all (J 0, J 1,..., J r ) LR(n, r), i J j where in Q r (b 1,..., b n ), ĉ i = c i for 1 i m and ĉ i = 0 for m + 1 i n.

8 Chi-Kwong Li, Yiu-Tung Poon, and Xuefeng Wang 8 If S = P r (a 1,..., a m ) Q r (b 1,..., b n ) is non-empty, then the minimal rank of ρ in S(σ 1, σ ) is r. Otherwise, the minimal rank is larger than r and we have to repeat Step with r increased by 1. By Theorem.1, Algorithm.9 will terminate for some r max{rank (σ 1 ), rank (σ )}. In fact, we can focus on σ 1 D m and σ D n with rank m and n, respectively. The set P r (a 1,..., a m ) and Q r (b 1,..., b n ) are polyhedral. There are standard linear programming packages for checking whether S = P r (a 1,..., a m ) Q r (b 1,..., b n ) is empty. Actually, we have the following. Proposition.10. Let S = P r (a 1,..., a m ) Q r (b 1,..., b n ) be defined as in Algorithm.9. Then the set S is non-empty if and only if a 1,..., a m, b 1,..., b n satisfy a finite set of linear inequalities. Proof. Because (a 1 /r,..., a m /r) P r (a 1,..., a m ), and P r (a 1,..., a m ) is governed by a finite set of inequalities, it is a non-empty polyhedron. Thus, there are finitely many extreme points expressed as linear combinations of a 1,..., a m. Now, Q r (b 1,..., b n ) is determined a finite set of inequalities, say, v t j x β j for j = 1,..., N, where v 1,..., v N R n with entries in {b 1,..., b n } and β 1,..., β N R. If all the extreme points of P r (a 1,..., a m ) lies in the complement of the half space defined by v t 1x β 1, then S is empty. Otherwise, the intersection of P r (a 1,..., a m ) and the half space defined by v t 1x β 1 is a non-empty polytope, and has a finite number of extreme points expressed as linear combinations of a 1,..., a m, b 1,..., b n. We can repeat the argument to this new polytope and the half space v t x β. We may conclude either the set and the half space has empty intersection or non-empty intersection. Repeating the process, we get a finite set of inequalities involving a 1,..., a m, b 1,..., b n, such that any one of them being violated will imply that S =, and S if all the inequalities are satisfied. By the above proposition, one can determine whether S = P r (a 1,..., a m ) Q r (b 1,..., b n ) by checking a finite set of inequalities in terms of a 1,..., a m, b 1,..., b n. Using this result, one may determine the set S r (m : σ ) = {σ D m : there is ρ S(σ, σ ) with rank at most r} for a given σ D n ; and S r (σ 1 : n) = {σ D n : there is ρ S(σ 1, σ) with rank at most r} for a given σ 1 D m. We have the following. Proposition.11. Suppose σ D n has eigenvalues b 1 b n. Then σ S r (m : σ ) if and only if σ has eigenvalues a 1 a m such that P r (a 1,..., a m ) Q r (b 1,..., b n ). Suppose σ 1 D m has eigenvalues a 1 a m. Then σ S r (σ 1 : n) if and only if σ has eigenvalues b 1 b m such that P r (a 1,..., a m ) Q r (b 1,..., b n ). Although one can determine whether the set S = P r (a 1,..., a m ) Q r (b 1,..., b n ) is non-empty by checking a finite set of inequalities, the number of inequalities involved may be very large. For low dimension

9 9 Ranks of Quantum States With Prescribed Reduced States case, the inequalities may be reduced to a smaller set after the redundant inequalities are removed. We illustrate this in the following proposition with a direct proof. It would be nice if one can give a description of non-redundant inequalities governing the eigenvalues of the reduced states of a bipartite state with prescribed rank. Given a = (a 1,..., a n ) and b = (b 1,..., b n ) R n, we say that a is majorized by b, denoted by a b, if for 1 k n 1 the sum of the k largest components of a is less than or equal to that of b. By Horn s result [3, a is the diagonal of some B H n with eigenvalues b 1,..., b n if and only if a b. Proposition.1. Suppose σ 1 D 3 has eigenvalues a 1 a a 3 and σ D 6 has eigenvalues b 1 b 6. Then σ 1 S (3 : σ ) if and only if a 1, a, a 3 satisfying 3 i=1 a i = 6 b j and the following inequalities: b 3 + b 6, b 4 + b 5 a 1 b 1 + b (.6) b 3 + b 4 + b 5 + b 6 a b 1 + b + b 3 + b 4 b 5 + b 6 a 3 b 1 + b 4, b + b 3. Proof. Suppose σ D 6 has eigenvalues b 1 b 6 and σ 1 S (3 : σ ) has eigenvalues a 1 a a 3. Then by Corollary.8, there exists a unitary matrix U M 6 such that U σ U = (C ij ) 1 i,j with λ(c 11 ) = λ(c ) = (c 1, c, c 3 ) and λ(c 11 + C ) = (a 1, a, a 3 ). Then there exist c 1 c c 3 and 3 3 unitary matrices V 1 and V such that diag (V i C iiv i ) = (c 1, c, c 3 ) for i = 1,. Hence, (c 1, c 1, c, c, c 3, c 3 ) (b 1,..., b 6 ) [3, and we have 1. a 1 c 1 b 1 + b.. a c 1 + c b 1 + b + b 3 + b a 3 c = (b 1 +b +b 3 +b 4 +b 5 +b 6 ) (c 1 +c 3 ) (b 1 +b +b 3 +b 4 +b 5 +b 6 ) (b +b 3 +b 5 +b 6 ) = b 1 +b 4, and (b 1 + b + b 3 + b 4 + b 5 + b 6 ) (b 1 + b 4 + b 5 + b 6 ) = b + b 3. Here, (c 1 + c 3 ) b + b 3 + b 5 + b 6, b 1 + b 4 + b 5 + b 6 follows from the fact that [ ({, 3, 6, 7}, {1, 3, 4, 5}, {1, 3, 4, 5}), ({1, 4, 5, 6}, {1, 3, 4, 5}, {1, 3, 4, 5}) LR(6, ). The other inequalities can be deduced by looking at µi 3 diag(a 1, a, a 3 ) and µi 6 diag(b 1,..., b 6 ), where µ = (b b 6 )/6. Given b 1 b 6, let S be the set of (a 1, a, a 3 ) satisfying a 1 a a 3, 3 i=1 a i = 6 i=1 b i = 6µ and (.6). Then S is a convex polyhedron in R 3. If S is non-empty, we can choose an extreme point (a 1, a, a 3 ) of S. Therefore, among the inequalities a 1 a a 3 and (.6), at least two equalities hold. If a 1 = a = a 3 = µ, then from (.6) we have b 3 + b 6, b 4 + b 5 µ b 1 + b 4, b + b 3, b 5 µ b 4 b 1 and b 6 µ b 3 b. Thus, diag(b 1, b 5 ) is unitarily similar to B 1 with diagonal entries µ b 4, b and diag(b, b 6 ) is unitarily similar to B with diagonal entries µ b 3, c; see [3. Thus, B = diag(b 1,..., b 6 ) is unitarily similar

10 Chi-Kwong Li, Yiu-Tung Poon, and Xuefeng Wang 10 to diag(b 3, b 4 ) B 1 B. There exists a permutation matrix P such that P BP = (D ij ) with D 11 = diag(b 3, µ b 4, b) and D = diag(µ b 3, b 4, c). By the trace condition, we see that b + c = µ. The result follows from Corollary.8. Suppose either a 1 > a or a > a 3. Then at least one of the equalities in (.6) holds. Consider the following cases: 1. a 1 = b 1 + b. Then we have b 3 + b 4 + b 5 + b 6 a = (b 3 + b 4 + b 5 + b 6 ) a 3 b 3 + b 4. Therefore, (a, a 3 ) (b 3 + b 4, b 5 + b 6 ). So there exists a unitary matrix U 1 such that U1 diag(b 3 + b 4, b 5 + b 6 )U 1 = (a, a 3 ). Let U = U 1 diag(1, 1). Let B 1 = 1 [ b3 + b 4 b 3 b 4 b 3 b 4 b 3 + b 4 and B = 1 [ b5 + b 6 b 5 b 6 b 5 b 6 b 5 + b 6 Then B = diag(b 1, b ) B 1 B has eigenvalues b 1,..., b 6. There exists a permutation matrix P such that P BP = (D ij ) with. D 11 = 1 diag(b 1, b 3 + b 4, b 5 + b 6 ) and D = 1 diag(b, b 3 + b 4, b 5 + b 6 ). Let U = P ([1 U 1 [1 U ). Then U BU will satisfy condition () in Corollary.8.. a = b 1 + b + b 3 + b 4. Then a 1 +a a = b 1 +b +b 3 +b 4 a 3 b 5 +b 6. Therefore, a 3 = b 5 +b 6 and (a 1, a ) (b 1 + b, b 3 + b 4 ). Thus, the result follows as in the previous case. 3. a 3 = b 1 + b 4. Then (b 1 + b + b 3 + b 4 + b 5 + b 6 ) = a 1 + a + a 3 3a 3 = 3(b 1 + b 4 ) (b 1 + b + b 3 + b 4 + b 5 + b 6 ) a 1 = a = a 3 = b 1 + b 4 = b + b 5 = b 3 + b 6. C = diag(b 1,..., b 6 ) will satisfy () in Corollary.8 4. a 3 = b + b 3. Then a 1 + a = b 1 + b 4 + b 5 + b 6 and a a 3 b 5 + b 6. Therefore, (a 1, a ) (b 1 + b 4, b 5 + b 6 ). Thus, the result follows as in the Case. The proof for the other equalities are similar. Note that the same set of inequalities (.6) will determine whether σ D 6 with eigenvalues b 1 b 6 lying in S (σ 1 : 6) for a given σ 1 D 3 with eigenvalues a 1 a a 3. In case a 3 = 0, then b 5 = b 6 = 0, and the set of inequalities reduce to: (b 3 + b 4 )/ a and a 1 b 1 + b. These inequalities will determine σ S (σ 1 : 4) with eigenvalues b 1 b 4 for a given σ 1 D with eigenvalues a 1 a. The same set of inequalities will also determine σ S ( : σ ) with eigenvalues a 1 a for a given σ D 4 with eigenvalues b 1 b 4.

11 11 Ranks of Quantum States With Prescribed Reduced States Note that a satisfies (.6) if and only if (c 1, c, c 3 ) a (b 1 + b, b 3 + b 4, b 5 + b 6 ), where ( b 4 + b 5, b 1 + b + b 3 + b 6, b ) 1 + b + b 3 + b 6 if 1 3 b 4 + b 5 c = ( 1 3, 1 3, 1 ) 3 ( b1 + b 4 + b 5 + b 6, b ) 1 + b 4 + b 5 + b 6, b + b 3 if b 4 + b b + b 3 if b + b Quantum channels. Recall that quantum channels are completely positive linear maps Φ : M m M n that admit the operators sum representation (3.7) φ(a) = F j AFj, for some n m matrices F 1,..., F j such that r F j F j = I m ; see [1, 6. By the result in [1, Φ is a quantum channel if and only if the Choi matrix C(Φ) = (Φ(E ij )) M m (M n ) is positive semidefinite and tr Φ(E ij ) = δ ij. Thus, the set of quantum channels can be identified with the set QC(m, n) = {P = (P ij ) M m (M n ) : P is positive semidefinite, (tr(p ij )) = I m } = {mρ D mn : tr (ρ) = I m /m}. Consequently, the set of quantum channels Φ : M m M n satisfying Φ(I m /m) = ρ D n can be identified with S(I m /m, ρ ). In particular, S(I n /n, I n /n) can be identified with the set of unital quantum channels from M n to M n. For a quantum channel Φ, its Choi rank is defined as the rank of its Choi matrix C(Φ). Moreover, it is known that Φ has Choi rank r if and only if r is the minimum number of matrices F 1,..., F r required in the operator sum representation of Φ. By Theorem.3, we have the following. Proposition 3.1. There is ρ S(I n /n, I n /n) of rank k if and only if 1 k n. Equivalently, there is a unital quantum channel with Choi rank k if and only if 1 k n. By the result in the previous section, we have the following. Proposition 3.. Let ρ D n. There is a quantum channel Φ : M m M n with Choi rank k and Φ(I m /m) = ρ if and only if there is a rank k element in S(I m /m, ρ ). As a result, the value k can be any value between the minimum value r determined by Algorithm.9 and the maximum value rank (ρ) m. Acknowledgments. We thank Dr. Cheng Guo for some helpful discussion. Also, we would like to thank the two referees for some helpful suggestions. Li is an honorary professor of the Shanghai University, and an affiliate member of the Institute for Quantum Computing, University of Waterloo; his research was supported by the USA NSF DMS , the Simons Foundation Grant , and NNSF of China Grant Part of the work was done while the Li and Poon were visiting the Institute for Quantum Computing at the University of Waterloo. They gratefully acknowledged the support and kind hospitality of the Institute.

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