Preservers of maximally entangled states

Preservers of maximally entangled states

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Linear Algebra and its Applications www.elsevier.com/locate/laa

Preservers of maximally entangled states Edward Poon Department of Mathematics, Embry–Riddle Aeronautical University, Prescott, AZ 86301, USA

a r t i c l e

i n f o

Article history: Received 30 July 2013 Accepted 6 March 2014 Available online xxxx Submitted by H. Woerdeman MSC: 15A86 15B57 46N50

a b s t r a c t The linear structure of the real space spanned by maximally entangled states is investigated, and used to completely characterize those linear maps preserving the set of maximally entangled states on Mm ⊗ Mm , where Mm denotes the space of m × m complex matrices. Aside from a degenerate rank one map, such preservers are generated by a change of orthonormal basis in each tensor factor, interchanging the two tensor factors, and the transpose operator. © 2014 Elsevier Inc. All rights reserved.

Keywords: Maximally entangled state Linear preserver Quantum state Density matrix

1. Introduction and notation Let Mm be the space of m×m complex matrices, and let Hm be the real space of m×m (complex) Hermitian matrices. On a finite-dimensional Hilbert space H of dimension m, a quantum state ρ is simply a density matrix in Hm (that is, ρ is a positive semi-definite m × m matrix of trace one). A state ρ is said to be pure if it has rank one (in other words, ρ is a rank one (orthogonal) projection). E-mail address: [email protected] http://dx.doi.org/10.1016/j.laa.2014.03.009 0024-3795/© 2014 Elsevier Inc. All rights reserved.

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In quantum information theory, one of the most important concepts is that of entanglement, which occurs when dealing with a multipartite system. We shall restrict our attention to one of the most common cases, a bipartite system H = HA ⊗ HB where HA and HB are Hilbert spaces of the same dimension m. In this case, a state ρ is said to be r separable if one can write ρ = i=1 pi ρi ⊗ σi for some states ρi , σi ∈ Hm and positive scalars pi summing to one. Otherwise we say the state is entangled. Entanglement is considered a valuable resource, responsible for the power of quantum computing (see [9] for a standard reference) and for applications such as superdense coding (see [3]) and quantum teleportation (see [2]). There are various ways to measure how much entanglement a state has (though for the bipartite case we consider, many of these measures are equivalent); those states possessing maximal entanglement are of particular importance, and a natural question is: what types of state transformations will preserve maximal entanglement? More generally, what linear transformations will map maximally entangled states back to themselves? The answer to this question forms the main result of this paper: Theorem 1.1. Let MES denote the set of maximally entangled states in Mm ⊗ Mm . A linear map Φ : Span(MES) → Span(MES) preserves MES if and only if it has one of the following forms: 1. A ⊗ B → (U ⊗ V )(A ⊗ B)σ (U ⊗ V )∗ for some unitary matrices U, V ∈ Mm . 2. A ⊗ B → (U ⊗ V )(B ⊗ A)σ (U ⊗ V )∗ for some unitary matrices U, V ∈ Mm . 3. X → (Tr X)ρ for some ρ ∈ MES. Here the map A → Aσ denotes either the identity or the transpose map. Questions of this type have a long history, and fall under the broader purview of linear preserver problems (two useful surveys are [6,7]). Recently there has been work done on finding and classifying linear preservers of various properties or sets related to quantum information theory. One paper of particular relevance is [4], in which the authors classify linear preservers of separable states (a related paper is [8]). However, they work under the more restrictive assumption that the linear map is surjective, an assumption we shall not require. Another paper which considers more specialized maps preserving maximal entanglement is [5]. This paper shall be organized as follows. Section 1 will conclude by introducing some notation. Section 2 will define what a maximally entangled state is, and investigate the real linear span of such states. Section 3 will prepare for the final section by stating and proving a number of technical lemmas. Finally, Section 4 will contain the proof of our main theorem, where we completely characterize the linear maps preserving maximally entangled states on Mm ⊗ Mm . We close this section by fixing some additional notation.

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We write Im and 0m for the m × m identity and zero matrices, respectively (omitting the subscript if the size is clear from the context). We let ei denote the (column) vector whose only nonzero entry is a 1 in the ith position and let Eij = ei e∗j . The group of n × n unitary matrices is denoted by Un . We shall abbreviate maximally entangled state(s) by MES, and will abuse notation by allowing MES to be both singular and plural, depending on the context (for example, we might say a state ρ is a MES, or we might refer to the set of all maximally entangled states as MES). The convex hull of MES is co(MES); the (real) linear span of MES is Span(MES). Given a bipartite system HA ⊗ HB , with dim HA = dim HB = m, the partial trace over system B is the linear map TrB : Mm ⊗ Mm → Mm defined by TrB (A ⊗ B) = (Tr B)A, where A ∈ Mm , B ∈ Mm , and Tr is the usual trace. The partial trace over the first system, TrA , is defined similarly. 2. Structure of maximally entangled states Let H = HA ⊗ HB be a bipartite system, where the Hilbert spaces HA and HB have the same dimension m. We identify HA and HB with Cm . Every pure state ρ on H may m be written as ρ = ψψ ∗ , with ψ = i=1 ci ui ⊗ vi for some nonnegative numbers ci and some orthonormal sets {u1 , . . . , um } and {v1 , . . . , vm } of Cm . This decomposition of ψ is known as the Schmidt decomposition; the Schmidt coefficients c1 , . . . , cm are unique. The entropy of entanglement in this case is given by S(TrA ρ) = S(TrB ρ) = −

m 

c2i log2 c2i ,

i=1

where S(A) = − Tr(A log2 A) is the von Neumann entropy. This entanglement measure is maximized when the Schmidt coefficients are all equal; since ψ is a unit vector, we have √ ci = 1/ m for all i when ρ is a pure MES. With this in mind, we make the following definitions. Definition 2.1. Let 1  ψ0 = √ ei ⊗ ei , m i=1 m

ρ0 = ψ0 ψ0∗ =

m 1  Eij ⊗ Eij , m i,j=1

and set 1  ψU,V = √ U ei ⊗ V e i , m i=1 m

∗ ρU,V = ψU,V ψU,V =

m 1  U Eij U ∗ ⊗ V Eij V ∗ . m i,j=1

Clearly MES = {ρU,V : U, V ∈ Mm are unitary}

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is the similarity orbit of ρ0 under the action of the unitary subgroup Um ⊗ Um , and hence is a compact, connected set of rank 1 projections. One easily sees that MES is invariant under local unitary rotations, transposition, and the ‘flip’ operator F , defined by F (A ⊗ B) = B ⊗ A. As such, it is not surprising that compositions of these three types of operations are linear preservers of MES. Of course, showing that these are the only ones (aside from trivial degenerate preservers) is significantly more difficult. 1 Note that if ρ is a MES, then TrA ρ = TrB ρ = m Im . In fact, this essentially characterizes the real linear span of MES. Let Sm = {X ∈ Hm ⊗ Hm : TrB X = TrA X = 0}; note this is a real vector space of dimension (m2 − 1)2 . Proposition 2.2. The real linear span of MES, denoted by Span(MES), is RIm2 + Sm . Proof. Note that if ρ is a MES, then ρ − m12 Im ⊗ Im lies in Sm . So it suffices to show that Im ⊗ Im and Sm lie in Span(MES). We shall repeatedly use the fact that MES, and hence Span(MES), is invariant under conjugation by U ⊗ V ∈ Um ⊗ Um . m If Wk = −2Ekk + i=1 Eii , then X=

ψ0 ψ0∗

− (W1 ⊗

In )ψ0 ψ0∗

 ∗  2 W1 ⊗ In = m



e1 e∗j



e1 e∗j

+

j=1



ei e∗1



ei e∗1



i=1

lies in Span(MES), and hence so does Y =

  m X − (W2 ⊗ In )X W2∗ ⊗ In = E12 ⊗ E12 + E21 ⊗ E21 . 4

Conjugating by Im ⊗ V for arbitrary unitary V ∈ Um shows that E12 ⊗ A + E21 ⊗ A∗ is in Span(MES) for any rank one nilpotent A of norm one. By linearity, E12 ⊗ A + E21 ⊗ A∗ is in Span(MES) for all A ∈ Mm with trace zero. Conjugating by U ⊗ Im for arbitrary permutations U ∈ Um shows that Eij ⊗ A + Eji ⊗ A∗ is in Span(MES) for all indices i = j, and all A ∈ Mm with trace zero. Then Z=

m 

   ∗ Eii ⊗ Eii = m ψ0 ψ0 − Eij ⊗ Eij

i=1

i=j

lies in Span(MES), and hence so does m   j=1

m    −j  Q ⊗ In Z Q ⊗ In = Im ⊗ Eii j

i=1

where Q ∈ Um is a cyclic permutation on m symbols. Thus Im ⊗ Im lies in Span(MES).

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Then   Z − (Im ⊗ Qk )Z Im ⊗ Q∗k = Ekk ⊗ (Ekk − Emm ) + Emm ⊗ (Emm − Ekk ) lies in Span(MES) for all k = 1, . . . , m − 1. Conjugating by Im ⊗ V for arbitrary V ∈ Um and using linearity shows that Ekk ⊗ B + Emm ⊗ (−B) ∈ Span(MES) for all k = 1, . . . , m − 1 and all B ∈ Hm with trace zero. By linearity, m−1 

Eii ⊗ Bi + Emm ⊗

i=1



m−1 

Bi

i=1

lies in Span(MES) for any B1 , . . . , Bm−1 ∈ Hm with trace zero. Thus Sm lies in Span(MES) as desired. 2 In general, the real linear span of an arbitrary set of projections contains additional projections not in the original set. This is not the case for MES. Remark 2.3. The set of pure states in the real linear span of MES is precisely MES. Proof. Suppose ψψ ∗ is a pure state in Span(MES). By the preceding proposition, 1 TrB ψψ ∗ = m Im . Using the Schmidt decomposition for ψ shows that ψψ ∗ ∈ MES as desired. 2 Lemma 2.4. Let U, V, W ∈ Um be unitary. Then ρU,V = ρI,W if and only if W = eiφ V U t for some φ ∈ R. Proof. Note ρU,V = ρI,W if and only if ψU,V = eiθ ψI,W for some θ ∈ R. Given a matrix A, let Ajk denote the (j, k)-entry of A. Then eiθ ψI,W = ψU,V  ∗     ⇐⇒ ej ⊗ e∗k eiθ ψI,W = e∗j ⊗ e∗k ψU,V ⇐⇒

eiθ Wkj =

⇐⇒

−iθ

m   i=1

W =e

VU

for all j = 1, . . . , m; k = 1, . . . , n

m      e∗j U ei e∗k V ei = Uji Vki = U V t jk i=1

t

as asserted. 2 Thus, every MES can be expressed as ρI,W for some unitary W , and clearly ρI,V = ρI,W if and only if W = eiφ V for some φ ∈ R. Note also that ρU,V = ρ0 if and only if U V t = eiθ I for some θ ∈ R. The following lemma gives some insight into the linear structure of MES.

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Lemma 2.5. Fix λ, μ ∈ (0, 1) and a unitary V1 ∈ Mm such that ρI,V1 = ρ0 . Then there exist unitaries V2 , V3 ∈ Mm satisfying λρ0 + (1 − λ)ρ1 = μρ2 + (1 − μ)ρ3

(1)

(here we write ρi = ρI,Vi for brevity) if and only if one of the following cases holds: 1. λ = μ, ρ0 = ρ2 , and ρ1 = ρ3 . 2. λ = 1 − μ, ρ0 = ρ3 , and ρ1 = ρ2 . 3. One can write V1 = eiω1 (cos θI + i sin θH) for some Hermitian unitary H = ±I and ω1 , θ ∈ [0, 2π), and the equation μei2α + (1 − μ)ei2β = λ + (1 − λ)ei2θ

(2)

is satisfied for some α, β ∈ [0, 2π). In such a case, the solutions are given by V2 = eiω2 (cos αI + i sin αH)

and

V3 = eiω3 (cos βI + i sin βH)

for any ω2 , ω3 ∈ [0, 2π). Proof. Suppose (1) holds. Write V1 ei = xi , V2 ei = yi , and V3 ei = zi , so {x1 , . . . , xm }, {y1 , . . . , ym }, {z1 , . . . , zm } are orthonormal bases of Cm . Since each term in Eq. (1) is positive semidefinite and of rank one, we must have that ψI,V2 , ψI,V3 both lie in the span of ψ0 , ψI,V1 . Hence m 

ei ⊗ yi = a

i=1

m 

e i ⊗ ei + b

i=1

m 

ei ⊗ xi

(3)

ei ⊗ x i

(4)

i=1

and m  i=1

ei ⊗ zi = c

m  i=1

ei ⊗ ei + δ

m  i=1

for some constants a, b, c, δ. Since replacing Vi with eiθi Vi does not affect Eq. (1), we may assume that a, b, c  0 and write δ = deiθ with d  0. Now if any of a, b, c, d are zero, then two of the states in (1) are identical (one from each side). The coefficients in front of these two states must match. (Otherwise, by re-arranging (1), we would obtain a linear combination of three pure states, with nonzero coefficients, equalling zero. This would imply that all the pure states ρi are identical, contradicting ρ1 = ρ0 .) Thus we

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must have either the first or second case, and it is clear that both cases give a solution to (1). Henceforth, we assume a, b, c, d > 0. Since V1 , V2 , V3 are unitary we have x∗i xj = δij

∀i, j = 1, . . . , m

and similarly yi∗ yj = δij , zi∗ zj = δij for all i, j = 1, . . . , m. Using yi = aei + bxi and zi = cei + deiθ xi , we see that these latter conditions are satisfied if and only if   1 − a2 − b2 Re e∗i xi = 2ab

∀i and e∗i xj = −x∗i ej

∀i = j

(5)

and   1 − c2 − d2 Re eiθ e∗i xi = 2cd

∀i and e2iθ e∗i xj = −x∗i ej

∀i = j.

(6)

Let k = (1 − a2 − b2 )/(2ab). We claim that V1 − kI is skew-Hermitian. Due to (5) it is clear that the claim holds if V1 is diagonal, so suppose V1 is not diagonal. Then for some index l, xl and el are linearly independent. Substituting (3) and (4) into (1) and simplifying shows that (1) holds if and only if   0 = ei e∗j μa2 + (1 − μ)c2 − λ + xi x∗j μb2 + (1 − μ)d2 − (1 − λ)   + ei x∗j μab + (1 − μ)cde−iθ + xi e∗j μab + (1 − μ)cdeiθ

(7)

for all i, j = 1, . . . , m. Since el e∗l , xl x∗l , el x∗l , xl e∗l are linearly independent, (7) holds if and only if the coefficients are all zero; in particular, eiθ must be real. By condition (6), (V1 )ij = −(V1 )ji for all i = j, i, j = 1, . . . , m; thus the claim holds. By (3) and (4), V2 = aI + bV1 and V3 = cI + δV1 , so each Vi is a linear combination of I and a skew-Hermitian matrix K = V1 − kI. Since each Vi is unitary, we must have V1 = cos θI + i sin θH, V2 = cos αI + i sin αH, V3 = cos βI + i sin βH for some Hermitian unitary H and real angles θ, α, β. Substituting these expressions into (1) and simplifying shows that (1) holds if and only if  0 = μ cos2 α + (1 − μ) cos2 β − λ − (1 − λ) cos2 θ Eij  + μ sin2 α + (1 − μ) sin2 β − (1 − λ) sin2 θ HEij H  + i/2 μ sin 2α + (1 − μ) sin 2β − (1 − λ) sin 2θ (HEij − Eij H)

(8)

for all i, j. We temporarily subdivide the proof into two cases. CASE I: Suppose H is diagonal. Note H = ±I (otherwise V1 would be a multiple of I and ρ1 = ρ0 , a contradiction), so Hek = ek and Hel = −el for some indices k, l. By considering the difference between (8) when i = k, j = l and (8) when i = l, j = k, we obtain

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μ sin 2α + (1 − μ) sin 2β = (1 − λ) sin 2θ.

(9)

Considering (8) when i = j = k and when i = k, j = l then gives μ cos2 α + (1 − μ) cos2 β = λ + (1 − λ) cos2 θ

(10)

μ sin2 α + (1 − μ) sin2 β = (1 − λ) sin2 θ.

(11)

and

CASE II: Suppose H is not diagonal, so Hel and el are linearly independent for some l. Then Ell , HEll H, and HEll − Ell H are linearly independent, so (8) holds if and only if (9), (10), and (11) do. Combining Case I and II: In both cases, we find that (1) holds if and only if (10) and (9) do (clearly (11) is equivalent to (10)). Using a half-angle formula, we may rewrite (10) as μ cos 2α + (1 − μ) cos 2β = λ + (1 − λ) cos 2θ. Adding this equation to i times (9), we see that (1) holds if and only if μei2α + (1 − μ)ei2β = λ + (1 − λ)ei2θ , so the lemma follows.

2

We shall refer to the solutions to (1) obtained in the first two cases of Lemma 2.5 as trivial solutions. In the nontrivial third case, note that condition (2) simply says that the line segment joining 1 and ei2θ on the unit circle intersects the line segment joining ei2α and ei2β , and divides each segment into fixed proportions λ and μ. From this geometric consideration, one can readily make the following observations. Remark 2.6. Eq. (1) has infinitely many solutions (for ρ2 and ρ3 ) if and only if λ = μ = 1/2 and V1 equals a complex unit times a Hermitian unitary. Somewhat more generally, if V1 is a skew-Hermitian unitary (so θ = π/2 in case 3 of Lemma 2.5), we have nontrivial solutions to (1) if and only if one of the following occur. 1. λ = 1/2 and μ = 1/2. In this case the solutions are ρ2 = ρI,cos φI+sin φV1 , ρ3 = ρI,sin φI−cos φV1 with φ arbitrary. 2. λ = 1/2 and μ lies strictly between λ and 1 − λ. In this case there are exactly two solutions: ρ2 = ρI,aI+bV1 and ρ3 = ρI,cI+dV1 , or ρ2 = ρI,aI−bV1 and ρ3 = ρI,cI−dV1 . Here a = cos α, b = sin α, c = cos β, d = sin β, where α, β give one particular solution to (2). Note abcd = 0.

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Lemma 2.5 shows that MES of the form ρI,U , with U skew-Hermitian, play a special role in the linear structure of Span(MES). A natural question is: does this smaller set of special MES span all of Span(MES)? The answer is no. First note that



T0 = ρI,U : U ∗ = −U is unitary = ρI,U : U ∗ = U is unitary since, by Lemma 2.4, there is freedom in choosing a complex phase for U . We claim that iE12 ⊗ E12 − iE21 ⊗ E21 lies in Span(MES) but does not lie in Span(T0 ). To see the latter assertion, suppose H is a Hermitian unitary. Then the entry in the E12 ⊗ E12 position of ρI,H

m 1  = Eij ⊗ HEij H ∗ m i,j=1

is e∗1 He1 e∗2 H ∗ e2 , which is always real. Hence the entry in the E12 ⊗ E12 position of any matrix in Span(T0 ) is real, so iE12 ⊗ E12 − iE21 ⊗ E21 ∈ / Span(T0 ). Also from Lemma 2.5, one observes that T = {ρI,U : U ∈ Ω}, where

Ω = eiθ (xI + iyH): H ∈ Hm ∩ Um ; x, y, θ ∈ R; x2 + y 2 = 1 , also plays a distinguished role in the linear structure of Span(MES). It requires a little more effort, but one can also show that iE12 ⊗ E12 − iE21 ⊗ E21 ∈ / Span(T ) either when m > 2, and so Span(T ) = Span(MES) if m > 2. (They are equal, however, when m = 2; this follows since Ω is precisely the set of unitaries with at most two distinct eigenvalues.) We are close, though; one can in fact extend T slightly to obtain a spanning set for Span(MES). Lemma 2.7. Let m > 2. The real linear span of  T+ = ρI,U : U ∈ Um has at most 2 distinct eigenvalues    0 or is unitarily similar to 0i −i ⊕ Im−2 is Span(MES). Proof. Let P1 , . . . , Pm ∈ Hm be orthogonal rank 1 projections. Set U = I − 2Pk , V = I − 2Pl , and W = I − 2Pk − 2Pl where k = l. Then ρI,U − ρ0 =

m 1  Eij ⊗ (−2Pk Eij − 2Eij Pk + 4Pk Eij Pk ); m i,j=1

summing over k, adding 4ρ0 , and then dividing by 4/m shows that m  i,j=1

Eij ⊗

m  k=1

Pk Eij Pk ∈ Span(T0 ) ⊂ Span(T+ ).

(12)

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We also have ρI,W − ρ0 − (ρI,U − ρ0 ) − (ρI,V − ρ0 ) =

m 4  Eij ⊗ (Pk Eij Pl + Pl Eij Pk ). m i,j=1

This shows that, for any orthogonal rank 1 projections P and Q, m 

Eij ⊗ (P Eij Q + QEij P ) ∈ Span(T0 ) ⊂ Span(T+ ).

(13)

i,j=1

ˆ we can use the spectral decomposition to write U ˆ = Given an arbitrary unitary U n k=1 αk Pm for some orthogonal rank 1 projections P1 , . . . , Pm with |αk | = 1 for all k. Note mρI,Uˆ =

m 

Eij ⊗

i,j=1

=

m 

m 

αk α ¯ l Pk Eij Pl

k,l=1

Eij ⊗

i,j=1

m 

Pk Eij Pk +

k=1

m  i,j=1

Eij ⊗



αk α ¯ l Pk Eij Pl .

k=l

We must show that ρI,Uˆ ∈ Span(T+ ). By (12), we see it suffices to show m 

  Eij ⊗ eiθ Pk Eij Pl + e−iθ Pl Eij Pk ∈ Span(T+ )

∀k = l;

i,j=1

by (13), it suffices to show m 

Eij ⊗ (iPk Eij Pl − iPl Eij Pk ) ∈ Span(T+ )

∀k = l.

(14)

i,j=1

Let U = I − 2Pk , V = I − 2Pl , and X = iPk − iPl + (I − Pk − Pl ). A computation gives 1 1 ρI,X − ρI,U − ρI,V 2 2 m   1 = Eij ⊗ i 2(Pl Eij Pk − Pk Eij Pl ) + (Pk Eij − Eij Pk ) − (Pl Eij − Eij Pl ) . m i,j=1 Since ρI,X , ρI,U , ρI,V all lie in T+ , the above equation shows that (14) will hold if m  i,j=1

for all rank 1 projections P .

Eij ⊗ i(P Eij − Eij P ) ∈ Span(T+ )

(15)

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Given any Hermitian unitary H and x, y > 0 satisfying x2 + y 2 = 1, xI + iyH is a unitary matrix with at most two distinct eigenvalues. Since ρI,xI+iyH = x2 ρ0 + y 2 ρI,H + xy

m   i  Eij ⊗ HEij − Eij H ∗ , m i,j=1

we have that i

m 

  Eij ⊗ HEij − Eij H ∗ ∈ Span(T+ )

i,j=1

for all Hermitian unitary H. Since every projection can be expressed as a linear combination of Hermitian unitaries, (15) holds for all projections P , and the proof is complete. 2 3. Technical lemmas In this section we state and prove a number of lemmas which will be needed to prove our main theorem in the next section. Our first result is a simple one; it will allow us to define a linear map on a quotient space in the proof of the main theorem. Lemma 3.1. For A ∈ Mm , AEij + Eij A∗ = 0 for all i, j = 1, . . . , m if and only if A = ikI for some k ∈ R. Proof. AEij + Eij A∗ = 0

∀i, j

⇐⇒ ⇐⇒

  e∗k AEij + Eij A∗ el = 0   Aki δjl + δik A∗ jl = 0.

∀i, j, k, l

Thus all off-diagonal entries of A are zero, and the diagonal entries Akk are all equal and pure imaginary, as claimed. 2 The next two results will allow us to extend a linear map on a quotient space of Hn to a very nice linear map on Hn . Lemma 3.2. Let r ∈ N. There do not exist four rank r projections Q1 , Q2 , Q3 , Q4 ∈ H2r such that, for any i = j, Qi + Qj equals a projection plus a real scalar multiple of I. Proof. Suppose Qi + Qj = Pij + kij I for some projections Pij and real numbers kij . Since Qi + Qj − kij I is a projection, 2 (1 − 2kij )(Qi + Qj ) + Qi Qj + Qj Qi + kij I = Qi + Qj − kij I.

Without loss of generality, we may assume that Q1 =

 Ir 0 . 0 0

(16)

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  k I 0 Case 1: Suppose k1j = 0, 1/2, or 1. Using (16), we deduce that Qj = 12 1j0 (k1j +1)I , which contradicts that Qj is a projection.  Case 2: Suppose k1j = 0 or 1. Using (16), we find that Qj = 00 I0r , since rank Qj = r. √   I 3Uj Case 3: Suppose k1j = 1/2. Using (16), we deduce that Qj has the form 14 √3U ∗ 3I j

for some unitary Uj ∈ Mr . Since we clearly cannot have Qi = Qj for i = j, it follows that Case 3 must occur for √   3Uj 1 √I at least one index j. Suppose k1i = 0 or 1 and k1j = 1/2. Then Qi + Qj = 4 3U ∗ 7I √

j

with multiplicity r, contradicting the fact that Qi + Qj must have has eigenvalues 1 ± eigenvalues of the form p and p + 1. Therefore, one must have k12 = k13 = k14 = 1/2. Similarly, one can show that k23 =k24 = k34 =√1/2. Then for 2  i < j  4, Qi + Qj − 3 2

0

3(U +U )

i j 1/2I is a projection of the form 14 √3(U +U )∗ . It follows that Ui + Uj = 0 4I i j for all 2  i < j  4. But this is impossible as U2 = −U3 = U4 = −U2 implies that U2 = 0. 2

˜ n denote the quotient space Hn /RI = {A + RI: A ∈ Hn }, and for Lemma 3.3. Let H ˜n → H ˜ n is a (real) linear map such that, for A ∈ Hn , let A˜ = A + RI. Suppose ψ˜ : H ˜ ˜ any projection P ∈ Hn , ψ(P ) contains a projection QP . Then there exists a linear map ˜ n is the canonical quotient ψ : Hn → Hn such that ψ˜ ◦ π = π ◦ ψ, where π : Hn → H map. Moreover, ψ has the form ψ(A) = U Aσ U ∗ for all A ∈ Hn , where U is a unitary matrix,  ∈ {−1, 0, 1}, and A → Aσ denotes either the identity or the transpose map. ˜ n containing more than one projection is ˜0, which Proof. Note that the only coset in H contains 0 and I. Let Pk denote the set of rank k projections in Hn , and let P˜k = π(Pk ). Since ψ˜ is continuous and P˜0 = P˜n , P˜1 , . . . , P˜n−1 form n disjoint path-connected com˜ P˜1 ) ⊆ P˜r for some r  0. If r = 0 then, since ψ˜ is linear and the set ponents, we have ψ( 0 and the lemma holds by taking ψ ≡ 0. of rank one projections span Hn , we have ψ˜ ≡ ˜ So suppose r > 0. We split the proof into two cases. Case 1: Suppose 1  r  n/2. Let P1 , . . . , Pn be orthogonal rank 1 projections, and ˜ P˜i ). For i = j, Pi + Pj is a projection, so let Qi be the unique projection (of rank r) in ψ( ˜ ˜ P˜j ) = Q ˜ ˜ ˜i + Q ˜ j ; that is, Qi +Qj +kij I  there exists a projection in ψ(Pi + Pj ) = ψ(Pi )+ ψ( is a projection for some kij ∈ R. If r < n/2, then rank(Qi +Qj ) < n, so 0 is an eigenvalue of Qi +Qj . For the spectrum of Q1 + Q2 + kI to lie in {0, 1}, we must have k = 0. Thus Qi + Qj is a projection, whence Qi Qj = Qj Qi = 0. Then Q1 , . . . , Qn are orthogonal rank r projections, so rn  n, whence r = 1. On the other hand, if 1 = r = n/2, then we have n  4, contradicting ˜ P˜1 ) ⊆ P˜1 . Lemma 3.2. Thus we must have r = 1, that is, ψ(  We now construct the map ψ. Given X ∈ Hn , we may write X = i ai Pi for some real  scalars ai and rank 1 projections Pi . Define ψ(X) = i ai QPi . Provided ψ is well-defined, this clearly defines a linear map, and

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(π ◦ ψ)(X) =



13

  ˜ X) ˜ P˜i ) = ψ˜ ˜ = (ψ˜ ◦ π)(X) ai ψ( ai P˜i = ψ(

i

i

since ψ˜ is linear. Thus, it suffices to show that ψ is well-defined.   We must show that if i ai Pi = 0, then i ai QPi = 0. Applying ψ˜ ◦ π to both sides  of 0 = i ai Pi gives ˜0 =



˜ P˜i ) = ai ψ(

i



˜P , ai Q i

i

   that is, i ai QPi ∈ RI. It follows that i ai QPi = 0 if and only if Tr( i ai QPi ) = 0. Thus, it suffices to show that Tr QP = Tr P = 1 for any rank 1 projection P . But this is ˜ P˜1 ) ⊆ P˜1 . true because ψ( Indeed, we also have ψ(P1 ) ⊆ P1 . By [1, Corollary 1], ψ must have one of the following forms: 1. ψ(A) = SAS ∗ for some S ∈ Mn , 2. ψ(A) = SAt S ∗ for some S ∈ Mn , or 3. ψ(A) = L(A)B for some linear functional L : Hn → R and B ∈ Hn of rank 1. We claim that the third case is inadmissible. Suppose, by way of contradiction, the third case occurs. We may assume that B is a rank 1 projection. Since ψ(P1 ) ⊆ P1 , we have L(A) = 1 for all rank one projections; since P1 spans Hn , we have L(A) = Tr A for all A ∈ Hn . But then ˜ ˜0) = ψ( ˜ I) ˜ = ψ˜ ◦ π(I) = π ◦ ψ(I) = nB, ˜ ˜0 = ψ( a contradiction. So, either the first or second case must occur. Since ψ(P ) is a rank one projection for any rank one projection P , S must be invertible. Since ψ(P )2 = ψ(P ) for all rank one projections P , we conclude that S must be unitary. Thus the lemma holds in this case. Case 2: Suppose n/2 < r < n. Then the map φ˜ = −ψ˜ satisfies the hypotheses of the ˜ P˜1 ) = −ψ( ˜ P˜1 ) ⊆ −P˜r = P˜n−r with 1  n − r < n/2. (Note that if P is theorem and φ( a projection, then −P˜ contains the projection I − P .) Thus Case 1 applies to φ˜ and the lemma holds. 2 The final two lemmas will allow us to single out a particular unitary as being essentially unique. Lemma 3.4. Let n  3 and T = {z ∈ C: |z| = 1}. Let  Rα =

cos α sin α

sin α − cos α



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and

R = Rα ⊕ V : α ∈ [0, 2π]; V ∈ Hn−2 ∩ Un−2 . Then  {U ∈ Un : RU ∈ THn ∀R ∈ R} = T

cos θ sin θ

  − sin θ ⊕ In−2 : θ ∈ [0, 2π] . cos θ

Proof. Suppose U is unitary and RU ∈ THn for all R ∈ R. Set  U0 =

  1 0 ⊕ In−2 U 0 −1



and write

A U= C

 B , D

where A ∈ M2 , D ∈ Mn−2 . We may remove an arbitrary complex phase by assuming that U0 is Hermitian. Suppose, by way of contradiction, that U0 has a nonzero off-diagonal entry in the kth row for some k  3. Let V = I −2ek−2 e∗k−2 . Note that a matrix X ∈ THn if and only if X ∗ = eiθ X for some θ ∈ R. Since 

  1 0 ⊕ V U = (I2 ⊕ V )U0 ∈ THn 0 −1

and U0 ∈ Hn , we must have U0 (I2 ⊕V ) = ((I2 ⊕V )U0 )∗ = −(I2 ⊕V )U0 . Thus, any entry of U0 not in the kth row or column is zero—but then U0 cannot be unitary, a contradiction. It follows that all off-diagonal entries of U0 not in the (1, 2)- or (2, 1)-positions are zero. Thus B = 0, C = 0, A is unitary, and D is diagonal with ±1 on the diagonal. Without loss of generality we may assume D11 = 1. Suppose, by way of contradiction, that Dkk = −1 for some k. Let V ∈ Hn−2 ∩ Un−2 be arbitrary. Since 

  1 0 ⊕ V U ∈ THn 0 −1

implies

V D ∈ THn−2 ,

we can use the same argument as before to conclude that P V P t is either block-diagonal or block-off-diagonal (where P is a permutation such that P DP t = Ir ⊕ (−In−2−r )). But V was arbitrary, so this is a contradiction. Thus D = In−2 . It follows that Rα A ∈ H2 for all α ∈ [0, 2π]. Considering α = 0 and α = π/2, we see that all four entries of A must be real. Direct computation reveals that 

cos θ A= sin θ for some θ ∈ [0, 2π], as asserted. 2

− sin θ cos θ



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15

Lemma 3.5. Let U ∈ Mn , n  3, be a unitary of the form  U=

cos θ sin θ

 − sin θ ⊕ In−2 cos θ

for some θ ∈ [0, 2π). Let 

0 A= −1

 1 ⊕ iIn−2 , 0

 B=

√1 2 − √12

√1 2 √1 2

 ⊕ eiπ/4 In−2 .

If AU and BU each have at most two distinct eigenvalues, then U = U = In .

 0 −1 1 0

⊕ In−2 or

Proof. The eigenvalues of AU are ei(π/2−θ) , e−i(π/2−θ) , i. Two of these three eigenvalues must be equal; considering all three cases we find that θ = 0, π/2, π, or 3π/2. When θ = π or 3π/2, BU has three distinct eigenvalues: eiπ/4 , ei3π/4 , e−i3π/4 . One can readily verify that, when θ = 0 or π/2, AU and BU have two distinct eigenvalues, so the lemma holds. 2 4. Linear preservers of maximally entangled states We would like to characterize the real linear maps on Hm ⊗ Hm which preserve the set of MES. Clearly there is a lot of freedom in devising such maps: if Φ is one such map, one ˜ can obtain another linear preserver Φ˜ by setting Φ(X) = Φ(X) for all X ∈ Span(MES), ˜ and defining Φ however we like on a basis for the orthogonal complement of Span(MES). Thus we should restrict our attention to real linear maps on Span(MES) which preserve the set of MES. Note that such a restriction makes this problem significantly more difficult: often one can solve preserver problems by making use of existing nice results for preservers on entire matrix spaces like Mm or Hm . In our case we cannot take recourse to such results, as there appears to be no natural way to extend a preserver on Span(MES) to a map on all of Hm ⊗ Hm . In practice, one may be concerned with affine maps on states which preserve MES. The following proposition shows that, to characterize affine maps of co(MES) preserving MES, it is sufficient to consider linear maps on Span(MES) preserving MES. Proposition 4.1. Let Φ : co(MES) → co(MES) be an affine map such that Φ(MES) ⊆ MES. Then Φ extends to a unique linear map Ψ : Span(MES) → Span(MES). Proof. Let K=

 k  i=1

 λi ρi : k ∈ N, λi  0, ρi ∈ MES

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16

be the cone of positive semidefinite matrices generated by MES. Since an element A ∈ K lies in co(MES) if and only if Tr A = 1, we can extend Φ to a map Ψ : K → K by defining Ψ (tρ) = tΦ(ρ) for any t  0 and any ρ ∈ co(MES). Clearly Ψ is affine and homogeneous. Given A ∈ Span(MES), we can write A = ρ+ − ρ− for some ρ+ , ρ− ∈ K. Set Ψ (A) = Ψ (ρ+ ) − Ψ (ρ− ). It is easy to check that this gives a well-defined linear map on Span(MES). It is clear that this extension is unique. 2 We now prove our main result. Proof of Theorem 1.1. Sufficiency is clear. To prove necessity, suppose Φ(MES) ⊆ MES. We may write Φ(ρ0 ) = (U ⊗ V )ρ0 (U ⊗ V )∗ for some unitaries U, V ∈ Mm . By replacing Φ with the map X → (U ⊗ V )∗ Φ(X)(U ⊗ V ) if necessary, we may assume Φ(ρ0 ) = ρ0 . Step 1. Let Pskew = {ρI,V : V ∈ Mm is skew-Hermitian unitary}. We claim that Φ(Pskew ) ⊆ Pskew . Let V1 be a skew-Hermitian unitary and write ρ1 = ρI,V1 . Suppose, by way of contra/ Pskew , so ρ1 = ρ0 . By Lemma 2.5 and Remark 2.6, the equation diction, that Φ(ρ1 ) ∈ 1 1 1 1 ρ0 + ρ1 = ρ2 + ρ3 2 2 2 2

(17)

has infinitely many solutions for MES ρ2 , ρ3 ; for example, one can take ρ2 = ρI,V2 where V2 = cos θI + sin θV1 , θ ∈ [0, 2π]. By applying Φ to both sides of (17), we see that the equation 1 1 1 1 ρ0 + Φ(ρ1 ) = ρ˜2 + ρ˜3 2 2 2 2

(18)

has solutions ρ˜2 = Φ(ρI,V2 ). But Φ(ρ1 ) ∈ / Pskew , so by Lemma 2.5, the only solutions to (18) are trivial, namely ρ˜2 = ρ0 or Φ(ρ1 ). Thus



Φ(ρI,cos θI+sin θV1 ): θ ∈ [0, 2π] ⊆ ρ0 , Φ(ρ1 ) .

Considering θ = 0, π/2, we see that these two sets are equal. But the set on the left-hand side is connected, while the set on the right-hand side is not, unless Φ(ρ1 ) = ρ0 . But this / Pskew . Thus our claim holds. contradicts Φ(ρ1 ) ∈ Step 2. We claim there is a map g from the set of Hermitian unitaries back to itself satisfying

m m     Φ Eij ⊗ (iHEij − iEij H) = Eij ⊗ ig(H)Eij − iEij g(H) i,j=1

i,j=1

and Φ(ρI,xI+iyH ) = ρI,xI+iyg(H) for all Hermitian unitaries H and all x, y ∈ R.

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17

Set g(I) = I and g(−I) = −I; since Φ(ρ0 ) = ρ0 , the asserted conditions hold. So now assume H = ±I is a Hermitian unitary. Write ρ1 = ρI,H , and suppose 0 < λ < μ < 1 − λ < 1 with μ = 1/2. By Lemma 2.5, λρ0 + (1 − λ)ρ1 = μρ2 + (1 − μ)ρ3

(∗)

is satisfied for ρ2 = ρI,aI+ibH , where a, b are as described in Remark 2.6. Since Φ preserves ˜ Applying Φ to (∗), we Pskew , we may write Φ(ρ1 ) = ρI,H˜ for some Hermitian unitary H. have λρ0 + (1 − λ)ρI,iH˜ = μΦ(ρ2 ) + (1 − μ)Φ(ρ3 ). Now if ρI,H˜ = ρ0 , then by Lemma 2.5 we must have Φ(ρ2 ) = ρI,aI+ibH˜ (in which case set ˜ or ρ ˜ g(H) = H) ˜ (in which case set g(H) = −H). On the other hand, if ρI,H ˜ = ρ0 I,aI−ibH ˜ = ±I. In this case set g(H) = I; since a2 + b2 = 1, then we must have Φ(ρ2 ) = ρ0 and H we have Φ(ρ2 ) = ρ0 = ρI,(a+ib)I = ρI,aI+ibg(H) . In any case, we have Φ(ρ2 ) = ρI,aI+ibg(H) . Now compare the two sides of this equation: m 

mρI,aI+ibg(H) = a2 ρ0 + b2 ρI,H˜ + ab

  Eij ⊗ ig(H)Eij − iEij g(H)

i,j=1

whereas mΦ(ρ2 ) equals

Φ

m 



Eij ⊗ (aI + ibH)Eij (aI + ibH)

i,j=1

2



m 

2

= Φ a ρ0 + b ρ1 + ab

Eij ⊗ (iHEij − iEij H)

i,j=1

= a2 ρ0 + b2 ρI,H˜ + abΦ

m 

Eij ⊗ (iHEij − iEij H) .

i,j=1

Since ab = 0, we have

Φ

m  i,j=1

Eij ⊗ (iHEij − iEij H)

=

m 

  Eij ⊗ ig(H)Eij − iEij g(H) ;

i,j=1

it then follows that Φ(ρI,xI+iyH ) = ρI,xI+iyg(H) for all x, y ∈ R, as claimed.

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Step 3: We claim that either (a) Φ(ρI,xI+iyH ) = ρ0 for all Hermitian unitary H and all x, y ∈ R satisfying x2 +y 2 = 1, or (b) there is a linear map ψ : Hm → Hm such that Φ(ρI,xI+iyH ) = ρI,xI+iyψ(H) for all Hermitian unitary H and all x, y ∈ R. Moreover, in this case ψ has the form ψ(A) = U Aσ U ∗ for all A ∈ Hm , where U is a unitary matrix,  ∈ {−1, 1}, and A → Aσ denotes either the identity or the transpose map. Since the real linear span of Hermitian unitaries is Hm , and since Φ is linear, Step 2 shows that for each Hermitian H we have

m m   ˆ ij − iEij H) ˆ Φ Eij ⊗ (iHEij − iEij H) = Eij ⊗ (iHE i,j=1

i,j=1

ˆ (depending on H). This allows us to define a real linear map for some Hermitian H ˜ ˜ ˜ ˜ m = Hm /RI = {A+RI: A ∈ Hm }) ˜ ψ : Hm → Hm (here Hm denotes the quotient space H ˜ X) ˜ ˜ m and define ψ( ˜ = Y˜ where as follows. Given X ∈ Hm , let X denote its coset in H

Φ

m  i,j=1

Eij ⊗ (iXEij − iEij X)

=

m 

Eij ⊗ (iY Eij − iEij Y ).

i,j=1

By Lemma 3.1, this map is well-defined, and it is clearly linear. Moreover, if H is a  where g is as in Step 2. Since P ∈ Hm is a ˜ H) ˜ = g(H), Hermitian unitary we have ψ( projection if and only if 2P − I is a Hermitian unitary, it follows that for any projection ˜ P˜ ) = Q  P there exists a projection QP such that ψ( P. By Lemma 3.3, there is a linear map ψ : Hm → Hm (of a particular form) satisfying  = g(H),  so that g(H) = ˜ ψ ◦ π = π ◦ ψ. Thus, for any Hermitian unitary H, ψ(H) ψ(H) + kH I for some kH ∈ R. Case (a): If ψ = 0 we have g(H) ∈ {±I} for all Hermitian unitary H. In this case Φ(ρI,xI+iyH ) = ρI,xI+iyg(H) = ρI,(x±iy)I = ρ0 for all Hermitian unitary H and x, y ∈ R satisfying x2 + y 2 = 1. Case (b): If ψ = 0, then ψ has the form ψ(A) = U Aσ U ∗ for all A ∈ Hm , where U is a unitary matrix,  ∈ {−1, 1}, and A → Aσ denotes either the identity or the transpose map. Thus any Hermitian unitary H = ±I is mapped to a Hermitian unitary ψ(H) = ±I. Since g(H) must be a Hermitian unitary it follows that kH = 0 for such H. Thus Φ(ρI,xI+iyH ) = ρI,xI+iyg(H) = ρI,xI+iyψ(H) for all Hermitian unitary H and all x, y ∈ R. The claim holds.

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19

Step 4: We normalize Φ in the following sense. Suppose case (b) in Step 3 holds. Then there is a linear map L on the real space RI + iHm such that Φ(ρI,xI+iyH ) = ρI,L(xI+iyH) for all Hermitian unitary H and all x, y ∈ R. More explicitly: if ψ(A) = U Aσ U ∗ then L(B) = U B σ U ∗ ; if ψ(A) = −U Aσ U ∗ then L(B) = U (B ∗ )σ U ∗ (here A → Aσ denotes either the identity or the transpose map). Thus L is a composition of unitary similarity, transpose, and/or complex conjugation. Each of these maps (acting on unitary matrices) give rise to maps on Span(MES) that are of the form asserted by Theorem 1.1; we list the correspondences below. Let f : Um → Um and define a map φ on MES by φ(ρI,A ) = ρI,f (A) . 1. Let f (A) = U A for some unitary U . Then φ(ρ) = (I ⊗ U )ρ(I ⊗ U ∗ ). 2. Let f (A) = AV for some unitary V . Then φ(ρI,A ) = ρI,AV = ρV t ,A by Lemma 2.4, so    ∗ φ(ρ) = V t ⊗ I ρ V t ⊗ I . ¯ Consider the map on Hm ⊗Hm given by B ⊗C → (B ⊗C)t = B t ⊗C t . 3. Let f (A) = A. Under this map, we have

ρU,V =

m m 1  1  ¯ U Eij U ∗ ⊗ V Eij V ∗ → U Eji U t ⊗ V¯ Eij V t = ρU¯ ,V¯ . m i,j=1 m i,j=1

Setting U = I we see that this is identical to the map φ(ρI,V ) = ρI,f (V ) . 4. Let f (A) = At . Consider the map U ⊗ V → V ⊗ U . Under this map, we have ρU,V =

m m 1  1  U Eij U ∗ ⊗ V Eij V ∗ → V Eij V ∗ ⊗ U Eij U ∗ = ρV,U = ρI,U V t m i,j=1 m i,j=1

by Lemma 2.4. Setting U = I we see that this is identical to the map φ(ρI,V ) = ρI,f (V ) . It follows that Φ is equal to an invertible map Ψ of the desired form when restricted to the set

T = ρI,xI+iyH : H ∈ Hm ∩ Um ; x, y ∈ R; x2 + y 2 = 1 .

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Since Φ is of the desired form (on all of Span(MES)) if and only if Ψ −1 ◦ Φ is, we may, without loss of generality, replace Φ with Ψ −1 ◦ Φ and assume that Φ fixes each element of T . Now, considering both cases from Step 3, we have either (a) Φ(ρ) = ρ0 for all ρ ∈ T , or (b) Φ(ρ) = ρ for all ρ ∈ T . It is worth recalling that T = {ρI,U : U ∈ Ω}, where

Ω = eiθ (xI + iyH): H ∈ Hm ∩ Um ; x, y, θ ∈ R; x2 + y 2 = 1 is precisely the set of unitaries with at most two distinct eigenvalues. Thus, for m = 2, we have T = MES. Since Φ is linear, Φ is either the identity map or the degenerate map X → (Tr X)ρ0 on Span(MES). In either case Φ is of the desired form and the theorem holds. Henceforth we assume m > 2. Step 5: With the normalization from Step 4 (namely, Φ(ρ) = ρ0 for all ρ ∈ T or Φ(ρ) = ρ for all ρ ∈ T ), we claim that either (a) Φ(ρ) = ρ0 for all ρ ∈ T+ , or (b) Φ(ρ) = ρ for all ρ ∈ T+ (note T+ is defined in Lemma 2.7). From this claim, combined with Lemma 2.7 and the linearity of Φ, the theorem follows immediately. Case A: Suppose we are in the case where Φ(ρ) = ρ0 for all ρ ∈ T . To show that Φ(ρ) = ρ0 for all ρ ∈ T+ , it suffices to show that Φ(ρI,Z ) = ρ0 where Z is any unitary matrix with eigenvalues i, −i, 1, . . . , 1 (counting multiplicity). By choosing a suitable orthonormal basis, we may write  Z=

 0 −1 ⊕ Im−2 . 1 0

Let  A=

0 −1

 1 ⊕ iIm−2 , 0

 B=

√1 2 − √12

√1 2 √1 2

 ⊕ eiπ/4 Im−2 .

With the aid of Lemma 2.5, one can verify that the equation λρ0 + (1 − λ)ρI,AZ = μρI,U2 + (1 − μ)ρI,U3

(19)

is satisfied for λ = 1/2 and μ = 1/6by U2 = aI + bW , U3 = cI + δW where W = √ √ −e−iπ/4 AZ, a = 1, b = 2, c = 2/5, and δ = −1/ 5. Applying the similarity transform X → (I ⊗ A∗ )X(I ⊗ A) to (19) gives

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λρI,A∗ + (1 − λ)ρI,Z = μρI,A∗ U2 + (1 − μ)ρI,A∗ U3 .

21

(20)

Applying Φ to (20), we get λρ0 + (1 − λ)Φ(ρI,Z ) = μρ0 + (1 − μ)ρ0 since A∗ ,

A∗ U2 =



 0 −i ⊕ (−Im−2 ), i 0

A∗ U3 =



0 3+i √ 10

3+i −√ 10 0



1 − 3i ⊕ √ Im−2 10

each have 2 distinct eigenvalues. Thus Φ(ρI,Z ) = ρ0 as desired. Case B: Suppose we are in the case where Φ(ρ) = ρ for all ρ ∈ T . To show that Φ(ρ) = ρ for all ρ ∈ T+ , it suffices to show that Φ(ρI,Z ) = ρI,Z where Z is any unitary matrix with eigenvalues i, −i, 1, . . . , 1 (counting multiplicity). As in Case A, we can choose a suitable orthonormal basis and write   0 −1 ⊕ Im−2 . Z= 1 0 We now make a few general observations. Consider the equation λρI,U0 + (1 − λ)ρI,U1 = μρI,U2 + (1 − μ)ρI,U3

(21)

where U0 , U1 , U2 , U3 ∈ Mm are unitary and λ, μ ∈ (0, 1). By applying the similarity transform X → (I ⊗ U0∗ )X(I ⊗ U0 ) to both sides of (21), we see that (21) holds if and only if λρ0 + (1 − λ)ρI,U0∗ U1 = μρI,U0∗ U2 + (1 − μ)ρI,U0∗ U3

(22)

does. We may write Φ(ρI,U ) = ρI,f (U ) for some function f : Um → Um . Then applying Φ to (21) gives λρI,f (U0 ) + (1 − λ)ρI,f (U1 ) = μρI,f (U2 ) + (1 − μ)ρI,f (U3 ) , which in turn holds if and only if λρ0 + (1 − λ)ρI,f (U0 )∗ f (U1 ) = μρI,f (U0 )∗ f (U2 ) + (1 − μ)ρI,f (U0 )∗ f (U3 )

(23)

does. From this we obtain two useful properties. 1. If U0∗ U1 has two distinct eigenvalues, then, by Lemma 2.5, (22) has solutions for unitary U2 , U3 when μ = 1/2 and λ = 1/4. It follows that (21) and hence (23) also have solutions when μ = 1/2 and λ = 1/4. By Lemma 2.5, f (U0 )∗ f (U1 ) has at most two distinct eigenvalues.

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E. Poon / Linear Algebra and its Applications ••• (••••) •••–•••

2. If U0∗ U1 is Hermitian with two distinct eigenvalues, then, by Lemma 2.5, (22) has infinitely many solutions for U2 , U3 when μ = λ = 1/2. In particular, U2 = (cos θ)U0 + (sin θ)U1 is a solution for any θ ∈ [0, π/2]. It follows that these choices for U2 will also give solutions to (23) when μ = λ = 1/2. Suppose for the moment that f (U0 )∗ f (U1 ) ∈ / THm , where T = {z ∈ C: |z| = 1}. By Lemma 2.5, (23) has only the trivial solutions ρI,f (U0 )∗ f (U2 ) = ρ0

or ρI,f (U0 )∗ f (U2 ) = ρI,f (U0 )∗ f (U1 ) .

But then

{ρ0 , ρI,f (U0 )∗ f (U1 ) } ⊆ ρI,f (U0 )∗ f (U2 ) : U2 = (cos θ)U0 + (sin θ)U1 , 0  θ  π/2 ⊆ {ρ0 , ρI,f (U0 )∗ f (U1 ) }; since the middle set is path connected, we have ρI,f (U0 )∗ f (U1 ) = ρ0 , whence f (U0 )∗ f (U1 ) ∈ TI, contradicting our assumption. Thus f (U0 )∗ f (U1 ) ∈ THm . By our assumption on Φ, if U ∈ Um has at most two distinct eigenvalues, f (U ) ∈ TU . If V ∈ Um and U V has at most two distinct eigenvalues then U f (V ) has at most two distinct eigenvalues by property (1). Similarly, if U V ∈ THm then U f (V ) ∈ THm by property (2). Using the notation  A=

0 −1

 1 ⊕ iIm−2 , 0

 B=

√1 2 − √12

√1 2 √1 2

 ⊕ eiπ/4 Im−2

and the results from Lemmas 3.4 and 3.5, it follows that {U ∈ Um : RU ∈ THm ∀R ∈ R, AU and BU have at most two distinct eigenvalues} = {TI, TZ}; on the other hand, if U lies in the set on the left-hand side, then so does f (U ). Thus f (Z) ∈ TI ∪ TZ, or equivalently, Φ(ρI,Z ) = ρ0 or ρI,Z . Suppose, by way of contradiction, that Φ(ρI,Z ) = ρ0 . Applying Φ to (20), we get λρI,A∗ + (1 − λ)ρ0 = μρI,A∗ U2 + (1 − μ)ρI,A∗ U3

(24)

by our assumption and since A∗ , A∗ U2 , A∗ U3 each have 2 distinct eigenvalues. But by Lemma 2.5, since A∗ is skew-Hermitian (resulting in θ = π/2 in case 3), there is no solution to (24) when λ = 1/2 and μ = 1/6. Hence Φ must fix ρI,Z and our proof is complete. 2 Note that if the preserver Φ is surjective, only the first two cases in Theorem 1.1 can occur. In particular, we have the following.

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E. Poon / Linear Algebra and its Applications ••• (••••) •••–•••

23

Corollary 4.2. Let Φ : Mm ⊗ Mm be an affine map such that Φ(co(MES)) = co(MES). Then Φ has one of the following forms, when restricted to Span(MES): 1. A ⊗ B → (U ⊗ V )(A ⊗ B)σ (U ⊗ V )∗ for some unitary matrices U, V ∈ Mm . 2. A ⊗ B → (U ⊗ V )(B ⊗ A)σ (U ⊗ V )∗ for some unitary matrices U, V ∈ Mm . Here the map A → Aσ denotes either the identity or the transpose map. Proof. Since the extreme points of the convex hull of MES is MES, we must have Φ(MES) = MES. By Proposition 4.1 and Theorem 1.1, the result follows. 2 A final comment on the case of a bipartite system Hm ⊗Hn where the subsystems have different dimensions m = n: we conjecture that the linear preservers in this case must have either form 1. or 3. from our main Theorem 1.1. Although the structural Lemma 2.5 can be extended to this case, our methods seem to break down around Step 3 of the proof. It would be interesting to verify or refute this conjecture. Acknowledgements The author would like to thank Professors Chi-Kwong Li and Yu Guo for some helpful discussions. Many thanks are also due to the referees for their suggestions to improve the paper; in particular, Lemma 3.2 and Lemma 2.5 were considerably simplified with the aid of one of the referees. References [1] E.M. Baruch, R. Loewy, Linear preservers on spaces of hermitian or real symmetric matrices, Linear Algebra Appl. 183 (1993) 89–102. [2] C.H. Bennett, G. Brassard, C. Crepeau, R. Jozsa, A. Peres, W. Wootters, Teleporting an unknown quantum state via dual classical and EPR channels, Phys. Rev. Lett. 70 (1993) 1895–1899. [3] C.H. Bennett, S.J. Wiesner, Communication via one- and two-particle operators on Einstein– Podolsky–Rosen states, Phys. Rev. Lett. 69 (1992) 2881–2884. [4] S. Friedland, C.-K. Li, Y.-T. Poon, N.-S. Sze, The automorphism group of separable states in quantum information theory, J. Math. Phys. 52 (2011) 042203. [5] Y. Guo, Z. Bai, S. Du, Local channels preserving maximal entanglement or Schmidt number, http://arxiv.org/pdf/1206.3119v2.pdf. [6] A. Guterman, C.-K. Li, P. Semrl, Some general techniques on linear preserver problems, Linear Algebra Appl. 315 (2000) 61–81. [7] C.-K. Li, S. Pierce, Linear preserver problems, Amer. Math. Monthly 108 (2001) 591–605. [8] C.-K. Li, Y.-T. Poon, N.-S. Sze, Linear preservers of tensor product of unitary orbits, and product numerical range, Linear Algebra Appl. 438 (2013) 3797–3803. [9] M. Nielsen, I. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2000.