Characterization of linear maps on M n whose multiplicity maps have maximal norm, with an application in quantum information
CCharacterization of linear maps on M n whose multiplicity maps have maximal norm,with an application in quantum information Daniel Puzzuoli
Department of Applied Mathematics and Institute for Quantum ComputingUniversity of Waterloo, Waterloo, Ontario, CanadaNovember 16, 2018
Given a linear map
Φ : M n → M m , its multiplicity maps are defined as thefamily of linear maps Φ ⊗ id k : M n ⊗ M k → M m ⊗ M k , where id k denotes theidentity on M k . Let k · k denote the trace-norm on matrices, as well as theinduced trace-norm on linear maps of matrices, i.e. k Φ k = max {k Φ( X ) k : X ∈ M n , k X k = 1 } . A fact of fundamental importance in both operatoralgebras and quantum information is that k Φ ⊗ id k k can grow with k . Ingeneral, the rate of growth is bounded by k Φ ⊗ id k k ≤ k k Φ k , and matrixtransposition is the canonical example of a map achieving this bound. Weprove that, up to an equivalence, the transpose is the unique map achieving thisbound. The equivalence is given in terms of complete trace-norm isometries,and the proof relies on a particular characterization of complete trace-normisometries regarding preservation of certain multiplication relations.We use this result to characterize the set of single-shot quantum channeldiscrimination games satisfying a norm relation that, operationally, impliesthat the game can be won with certainty using entanglement, but is hard towin without entanglement. Specifically, we show that the well-known exampleof such a game, involving the Werner-Holevo channels, is essentially the uniquegame satisfying this norm relation. This constitutes a step towards a charac-terization of single-shot quantum channel discrimination games with maximalgap between optimal performance of entangled and unentangled strategies. For a linear map
Φ : M n → M m , it is a well-known phenomenon that the norm of themultiplicity maps Φ ⊗ id k : M n ⊗ M k → M m ⊗ M k can grow with k . This phenomenonhas been extensively studied within the theory of C ∗ -algebras, leading to the topic of com-pletely bounded maps [1–3]. Within the field of quantum information, this phenomenonis connected to the study of entanglement. For a density matrix ρ ∈ M n ⊗ M k , if k (Φ ⊗ id k )( ρ ) k > k Φ k , (1) then ρ is entangled, and a well-known result in quantum information is that the existenceof a positive linear map Φ for which the above holds is also necessary for ρ to be entangled Accepted in
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Quantum2018-01-16, click title to verify a r X i v : . [ qu a n t - ph ] F e b negativ-ity [5], which, up to additive and multiplicative scalars is defined as (cid:13)(cid:13) ( T n ⊗ id k )( ρ ) (cid:13)(cid:13) , where T n is the transpose on M n . Hence, the growth of the norms of multiplicity maps is of fun-damental mathematical interest, and is deeply connected to the study and quantificationof entanglement in quantum information. While the norm of Φ ⊗ id k may grow with k , the growth rate is limited. For any linearmap Φ : M n → M m it generically holds that k Φ ⊗ id k k ≤ k k Φ k . (2) In this paper we are concerned with linear maps on matrices and will use the trace-norm,but we note that this fact is known much more generally (in terms of the C ∗ -norm) formaps on unital C ∗ -algebras (see [3, Exercise 3.10]). Provided k ≤ n , the canonical exampleof a map achieving equality in Equation (2) is the matrix transpose T n [6].Our main result is a characterization of the maps achieving equality in Equation (2) .We prove that, up to an equivalence, the transpose is in fact the only map achievingequality in Equation (2) . More specifically, a linear map Φ : M n → M m satisfies Equation (2) with equality if and only if there exists an isometric embedding of M k into M n onwhich Φ acts as the transpose followed by a complete trace-norm isometry.The proof relies on a characterization of complete trace-norm isometries particularlysuited to the problem. This characterization (among others proved) relates to how completetrace-norm isometries preserve certain multiplication relations. For example, if a linearmap Φ : M n → M m is a complete trace-norm isometry, and if A ∗ B = C ∗ D for A, B, C, D ∈ M n , then Φ( A ) ∗ Φ( B ) = Φ( C ) ∗ Φ( D ) . These statements and their proofs are somewhatsimilar to multiplicative domain proofs for unital completely-positive maps on C ∗ -algebras[7] (see also [3, Theorem 3.18]). We remark that the structure of complete trace-normisometries on M n , and consequently some of the other characterizations we give, maybe deduced from the more general structure of (not necessarily complete) trace-normisometries given in [8]. Nevertheless, we give self-contained proofs, and in some cases areable to utilize the “complete” assumption to prove certain implications in more generality(e.g. when the domain is a subspace V ⊂ M n ), which may be of independent interest.We also apply the main result in the setting of single-shot quantum channel discrimi-nation, which is the task of determining which of two known quantum channels is actingon a system given only a single use of the channel. As we will describe in detail in Sec-tion 5, this task may be formulated as a game parametrized by a triple ( λ, Γ , Γ ) , where Γ , Γ : M n → M m are quantum channels, and λ ∈ [0 , is a probability. Letting ||| · ||| denote the completely bounded trace-norm (see Section 2), we characterize such triplessatisfying the norm relation ||| λ Γ − (1 − λ )Γ ||| = n k λ Γ − (1 − λ )Γ k . (3) Operationally, the above norm relations imply that the game can be won with certaintyusing entanglement, but is hard to win without entanglement. In particular, we prove thatthe triple ( λ, Γ , Γ ) satisfies Equation (3) if and only if it is in some sense equivalent toa game involving the Werner-Holevo channels which is known to satisfy Equation (3) (see[9, Example 3.36]).In Section 2 we provide some background and definitions. In Section 3 we prove variouscharacterizations of complete trace-norm isometries. One characterization in particular is Note that, in quantum information the trace-norm is typically used, whereas in operator algebras theoperator norm is typically used. Due to the duality of these norms, it is possible to translate facts aboutone norm into facts about the other. For example, in this paper, we reference the C ∗ -algebra literature forfacts about the trace-norm, even though the trace-norm does not explicitly appear in the references. Forreaders unfamiliar with this duality, we describe the relevant facts in Appendix A. Accepted in
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Quantum2018-01-16, click title to verify pecially suited for later use, but we also include other characterizations; including onethat may naturally be interpreted as a linear map being a complete trace-norm isometry ifand only if its Choi matrix is maximally entangled. In Section 4 we prove the main result,which characterizes the linear maps Φ : M n → M m for which k Φ ⊗ id k k = k k Φ k , andas a Corollary characterize the maps for which ||| Φ ||| = n k Φ k . In Section 5 we use themain result to prove that the Werner-Holevo channel discrimination game is in some sensethe unique game (with input dimension n ) satisfying Equation (3) . Finally, we end witha discussion of open problems in Section 6. For an integer ≤ a ≤ n we denote e a ∈ C n to be the vector with a in the a th entry,and zeroes everywhere else. Similarly, for integers ≤ a, b ≤ n , we denote E a,b ∈ M n tobe the elementary matrix with a in the ( a, b ) -entry and zeroes in all other entries.For a matrix A , the trace-norm is defined as k A k = Tr( √ A ∗ A ) , and for a linear map Φ : M n → M m , the induced trace-norm of Φ is given by k Φ k = max {k Φ( X ) k : X ∈ M n , k X k ≤ } . (4) The completely bounded trace norm is given by ||| Φ ||| = sup (cid:8) k Φ ⊗ id k k : k ≥ (cid:9) = k Φ ⊗ id n k . (5) For a linear map
Φ : M n → M m , we will use J (Φ) to denote its Choi matrix [10], whichwe define as J (Φ) = n X a,b =1 Φ( E a,b ) ⊗ E a,b . (6) We will use T n to denote the transpose on M n , and write T n ( A ) or A T to denote thetranspose of a matrix A ∈ M n .We will need a few concepts from quantum information, even for the sections notdirectly involving that topic. An element ρ ∈ M n is called a density matrix if ρ ≥ and Tr( ρ ) = 1 . A quantum channel is a linear map Γ : M n → M m that is completely positiveand trace preserving.We will also use the term maximal entanglement . A unit vector u ∈ C n ⊗ C m is called maximally entangled if, for r = min( n, m ) , there exists orthonormal sets of unit vectors { x a } ra =1 ⊂ C n and { y a } ra =1 ⊂ C m for which u = r r r X a =1 x a ⊗ y a . (7) As mentioned in the introduction, the negativity of a density matrix ρ ∈ M n ⊗ M m isdefined (up to multiplicative and additive scalars) as k ( T n ⊗ id m )( ρ ) k [5]. This expressionis meant to quantify the entanglement of the density matrix ρ . If ρ is pure , i.e. ρ = uu ∗ for a unit vector u ∈ C n ⊗ C m , then the negativity achieves its maximum value of min( n, m ) if and only if u is maximally entangled. Theorem 7 of [11], which we now state,provides a characterization of matrices X ∈ M n ⊗ M m with k X k = 1 and satisfying k ( T n ⊗ id m )( X ) k = n . While it is not physically meaningful if X is not a density matrix,the theorem loosely provides a notion of “maximal entanglement” for arbitrary elementsof M n ⊗ M m . Accepted in
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Quantum2018-01-16, click title to verify heorem 1. Let X ∈ M n ⊗ M m with k X k ≤ . The following are equivalent.1. k ( T n ⊗ id m )( X ) k = n .2. m ≥ n , and there exists a positive integer r ≤ m/n , a density matrix σ ∈ M r , andisometries U, V : C n ⊗ C r → C m for which X = ( n ⊗ U )( τ n ⊗ σ )( n ⊗ V ∗ ) , (8) where τ n = n P na,b =1 E a,b ⊗ E a,b ∈ M n ⊗ M n is the canonical maximally entangledstate.If X is a density matrix then the above equivalence holds with V = U . Remark 2.
We will make use of the additional special case of the above theorem when X is Hermitian. In this case the second statement may be rewritten as: m ≥ n , and thereexists a positive integer r ≤ m/n , a Hermitian matrix H ∈ M r with k H k = 1, and anisometry U : C n ⊗ C r → C m for which X = ( n ⊗ U )( τ n ⊗ H )( n ⊗ U ∗ ) . (9)The only change necessary to the proof is to take a spectral decomposition of X , ratherthan a singular value decomposition. The rest of the proof follows as before. The last notion in quantum information that we will make use of is that of reversiblequantum channels. A linear map
Φ : M n → M m is called a reversible quantum channel if itis a quantum channel, and has a left inverse Ψ : M m → M n that is also a quantum channel. Due to its connection to error correction, conditions for reversibility (or recoverability) ofa channel continue to be extensively studied in various settings (for example, see [12, 13]).
Let V ⊂ M n be a subspace, and Φ : V → M m be a linear map. We say that Φ is a k -trace-norm isometry (or that it is k -trace-norm isometric ) if Φ ⊗ id k : V ⊗ M k → M m ⊗ M k is a trace-norm isometry, and say that it is a complete trace-norm isometry (or that it is completely trace-norm isometric ) if it is a k -trace-norm isometry for all integers k ≥ .The purpose of this section is to give various characterizations of complete trace-norm isometries taking M n into M m . Note that the structure of surjective operator normisometries (and hence surjective complete operator norm isometries) between C ∗ -algebrasis well-known [14]. Furthermore, in the matrix algebra case, a characterization of (notnecessarily surjective) operator norm isometries mapping M n → M k has been given forthe case k ≤ n − [15]. However, the dual/adjoint of a trace-norm isometry need not bean operator norm isometry, and so it is not possible to import those results here.We give the various characterizations in the theorem below. Remarks and some back-ground on what is already known, as well as some intermediate results, are given beforeits proof. This terminology is motivated by the fact that quantum channels model physical processes, and soa quantum channel having a left inverse that is also a quantum channel means that it can be physicallyundone, or reversed.
Accepted in
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Quantum2018-01-16, click title to verify heorem 3. For a linear map
Φ : M n → M m the following are equivalent.1. Φ is a complete trace-norm isometry.2. Φ is a 2-trace-norm isometry.3. For A, B, C, D ∈ M n the following implications hold: • A ∗ B = C ∗ D = ⇒ Φ( A ) ∗ Φ( B ) = Φ( C ) ∗ Φ( D ) , and • AB ∗ = CD ∗ = ⇒ Φ( A )Φ( B ) ∗ = Φ( C )Φ( D ) ∗ ,and k Φ( X ) k = k X k for some X ∈ M n \ { } .4. For A, B ∈ M n the following implications hold: • A ∗ B = 0 = ⇒ Φ( A ) ∗ Φ( B ) = 0 , and • AB ∗ = 0 = ⇒ Φ( A )Φ( B ) ∗ = 0 ,and k Φ( X ) k = k X k for some X ∈ M n \ { } .5. For rank-1 A, B ∈ M n the following implications hold: • A ∗ B = 0 and A ∗ A = B ∗ B = ⇒ Φ( A ) ∗ Φ( B ) = 0 , and • AB ∗ = 0 and AA ∗ = BB ∗ = ⇒ Φ( A )Φ( B ) ∗ = 0 ,and k Φ( X ) k = k X k for some X ∈ M n \ { } .6. k J (Φ) k = n and k J (Φ T n ) k = n .7. m ≥ n , and there exists a positive integer r ≤ m/n , a density matrix σ ∈ M r , andisometries U, V : C n ⊗ C r → C m for which Φ( X ) = U ( X ⊗ σ ) V ∗ (10) for all X ∈ M n .8. ||| Φ ||| = 1 , and Φ has a left inverse Ψ : M m → M n with ||| Ψ ||| = 1 .If, in addition, Φ is positive, then statement 7 holds with V = U , making Φ a quantumchannel, and Ψ may also be taken to be a quantum channel in statement 8 (and hence, Φ is a reversible quantum channel). Before continuing some comments on the theorem are in order. In statement 6, the norm k J (Φ T n ) k appears, but this specific location of the transpose is an arbitrary notationalchoice. Using the definition of the Choi matrix and properties of the transpose, it may beverified that k J (Φ T n ) k = k J ( T m Φ) k = k ( T m ⊗ id n )( J (Φ)) k = k (id m ⊗ T n )( J (Φ)) k (11) for any linear map Φ : M n → M m . As per the discussion in Section 2 regarding entangle-ment negativity, the expressions in Equation (11) can be roughly interpreted as a measureof how entangled J (Φ) is. With this interpretation, the characterization given in statement6 says that Φ is a complete trace-norm isometry if and only if its Choi matrix is maximallyentangled (and has a particular normalization).Statements 3, 4, and 5 concern the map Φ preserving certain kinds of multiplication.The intuition for these statements comes from the explicit structure given in statement Accepted in
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Quantum2018-01-16, click title to verify . However, in our proof, we show how they follow directly from the assumptions of Φ being either a complete or 2-trace-norm isometry. The benefit of these alternative proofs,which we give separately in Proposition 5 below, is that they work in more generality (i.e.in the proposition we only use that the domain is a subspace V ⊂ M n ), and so may beof independent interest. We also note that statement 5 may seem oddly specific, but it isincluded for being specially suited for proving the theorem in the following section.Lastly, while we give a complete proof of the above theorem, several equivalences maybe deduced from [8], whose title “Isometries for Ky Fan Norms between Matrix Spaces”is self-explanatory of its content. In particular, as a special case of the results therein,an explicit structural characterization of (not necessarily complete) trace-norm isometriestaking M n into M m is given. From this, the explicit structure of complete trace-normisometries may be deduced by refining this structure, and indeed, this refinement onlyrequires the additional assumption that the map is a -trace-norm isometry. Thus, theequivalence of statements 1, 2, and 7 may be viewed as a special case of the main theoremin [8]. Furthermore, the general technique of the proofs we give are in line with those of [8],and with linear norm preserver problems more generally [16]: translating between normrelations and algebraic relations for matrices. (See [17] for a survey of results on isometriesof matrix spaces for unitarily invariant norms.)With this last comment, we begin the proof of Theorem 3 with the following equivalencebetween a trace-norm relation for a × block-matrix, and statements about how theblocks multiply. Proposition 4.
For matrices
A, B, C, D ∈ M n , it holds that (cid:13)(cid:13)(cid:13)(cid:13) A BC D !(cid:13)(cid:13)(cid:13)(cid:13) = k A k + k B k + k C k + k D k , (12) if and only if A ∗ B = AC ∗ = D ∗ C = DB ∗ = 0 . (13) Proof.
In [11, Proposition 6] it was shown that, for any Hilbert-Schmidt orthogonal set ofmatrices { A i } ri =1 – all with the same dimensions, not necessarily square – it holds that (cid:13)(cid:13)(cid:13)(cid:13) r X i =1 A i (cid:13)(cid:13)(cid:13)(cid:13) = r X i =1 k A i k , (14)if and only if A ∗ i A j = 0 and A i A ∗ j = 0 for all i = j . The current proposition follows byapplication of this fact to the set { A ⊗ E , , B ⊗ E , , C ⊗ E , , D ⊗ E , } ⊂ M n ⊗ M . Next, we prove a proposition containing some of the implications required for Theorem3, but in more generality. We note that the proof takes inspiration from multiplicativedomain proofs for unital and completely positive linear maps on C ∗ -algebras (see [7] and[3, Theorem 3.18]). Proposition 5.
Let V ⊂ M n be a subspace, and let Φ : V → M n be linear.1. If Φ is a 2-trace-norm isometry, then for A, B ∈ V the following implications hold: • A ∗ B = 0 = ⇒ Φ( A ) ∗ Φ( B ) = 0 , and • AB ∗ = 0 = ⇒ Φ( A )Φ( B ) ∗ = 0 .2. If Φ is a complete trace-norm isometry, then for A, B, C, D ∈ V the following im-plications hold: Accepted in
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Quantum2018-01-16, click title to verify A ∗ B = C ∗ D = ⇒ Φ( A ) ∗ Φ( B ) = Φ( C ) ∗ Φ( D ) , and • AB ∗ = CD ∗ = ⇒ Φ( A )Φ( B ) ∗ = Φ( C )Φ( D ) ∗ .Proof. First, assume Φ is a 2-trace-norm isometry and let
A, B ∈ V . Assuming A ∗ B = 0,we have (cid:13)(cid:13)(cid:13)(cid:13) Φ( A ) Φ( B )0 0 !(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13) A B !(cid:13)(cid:13)(cid:13)(cid:13) = k A k + k B k = k Φ( A ) k + k Φ( B ) k , (15)where the second equality is by Proposition 4. Hence, also by Proposition 4, equalitybetween the first and last expressions implies that Φ( A ) ∗ Φ( B ) = 0. Similarly, if AB ∗ = 0,then (cid:13)(cid:13)(cid:13)(cid:13) Φ( A ) 0Φ( B ) 0 !(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13) A B !(cid:13)(cid:13)(cid:13)(cid:13) = k A k + k B k = k Φ( A ) k + k Φ( B ) k , (16)and so Φ( A )Φ( B ) ∗ = 0.Next, assume Φ is a complete trace-norm isometry on V , and let A, B, C, D ∈ V . If A ∗ B = C ∗ D , then A − C ! ∗ B D ! = 0 . (17)Under the assumption that Φ is completely trace-norm isometric, Φ ⊗ id is a 2-trace-normisometry, and so, by the 2-trace-norm isometry case, it holds that Φ( A ) 0 − Φ( C ) 0 ! ∗ Φ( B ) 0Φ( D ) 0 ! = 0 , (18)giving Φ( A ) ∗ Φ( B ) = Φ( C ) ∗ Φ( D ). Similarly, if AB ∗ = CD ∗ , then A − C ! B D ! ∗ = 0 , (19)and again the 2-trace-norm isometry case implies that Φ( A ) − Φ( C )0 0 ! Φ( B ) Φ( D )0 0 ! ∗ = 0 , (20)giving Φ( A )Φ( B ) ∗ = Φ( C )Φ( D ) ∗ . Proof of Theorem 3.
We prove the implications in the diagram below.1 2 34567 8
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Quantum2018-01-16, click title to verify he implications appear in the order: 1 ⇒
2, 3 ⇒ ⇒
5, 1 ⇒
6, 2 ⇒
4, 1 ⇒
3, 8 ⇒
1, 6 ⇒
7, 7 ⇒
8, and 5 ⇒
7. All implications except 5 ⇒
7, which is technically involved,follow essentially immediately from facts already given. The modified statements for thespecial case when Φ is positive are given before the proof of 5 ⇒ ⇒ ⇒ ⇒
5. The implication 1 ⇒ k J (id n ) k = n and k J ( T n ) k = n , (21)and Φ being a complete trace-norm isometry. The implications 2 ⇒ ⇒ ⇒ ⇒
7, the norm values of statement 6 imply by Theorem 1 that there exists apositive integer r ≤ m/n , a density matrix σ ∈ M r , and isometries U, V : C r ⊗ C n → C m for which 1 n J (Φ) = ( U ⊗ n )( σ ⊗ τ n )( V ∗ ⊗ n ) . (22)This is equivalent to the required form for Φ.For 7 ⇒
8, define Ψ( Y ) = Tr M r ( U ∗ Y V ) for Y ∈ M m , where Tr M r is the partial traceof M r . Using the fact that the trace-norm is non-increasing under partial trace, it may beverified that Ψ has the required properties.For the special case when Φ is positive, return to the proof of the implication 6 ⇒
7. AsΦ is Hermiticity preserving, by Remark 2 and Theorem 1, there exists a Hermitian H ∈ M r with k H k = 1, and an isometry U : C n ⊗ C r → C m for which Φ( X ) = U ( X ⊗ H ) U ∗ forall X ∈ M n . That Φ is positive implies H ≥
0, making H a density matrix, and giving Φthe required form. To see that Ψ may also be taken to be a quantum channel in statement8, we define Ψ as as in the proof of 7 ⇒ η ∈ M n , and set Ψ( Y ) = Tr M r ( U ∗ Y U ) + Tr(( m − U U ∗ ) Y ) η for all Y ∈ M m . It is routineto verify that Ψ is a quantum channel and that ΨΦ = id n .Lastly, we show 5 ⇒
7. We will use the assumption in statement 5 to build furtherfacts about how outputs of Φ on rank-1 matrices multiply, which we break into a series ofclaims.
Claim 1.
For unit vectors x , x , y ∈ C n with h x , x i = 0, it holds thatΦ( x y ∗ ) ∗ Φ( x y ∗ ) = Φ( x y ∗ ) ∗ Φ( x y ∗ ) and Φ( yx ∗ )Φ( yx ∗ ) ∗ = Φ( yx ∗ )Φ( yx ∗ ) ∗ . (23)To see the first equality, note that x + x ⊥ x − x , and so0 = Φ(( x + x ) y ∗ ) ∗ Φ(( x − x ) y ∗ ) (24)= Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) + Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) (25)= Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) , (26)where the second and third term in Equation (25) are 0 by application of statement 5. Thisgives the desired equality, and it follows similarly that Φ( yx ∗ )Φ( yx ∗ ) ∗ = Φ( yx ∗ )Φ( yx ∗ ) ∗ . Claim 2.
For any x , x , y , y ∈ C n with h x , x i = h y , y i = 0, it holds thatΦ( x y ∗ ) ∗ Φ( x y ∗ ) = 0 and Φ( x y ∗ )Φ( x y ∗ ) ∗ = 0 . (27)For the first equality, assuming without loss of generality that x , x , y , and y have unitlength, Claim 1 gives that Φ( x y ∗ )Φ( x y ∗ ) ∗ = Φ( x y ∗ )Φ( x y ∗ ) ∗ , and statement 5 givesthat Φ( x y ∗ ) ∗ Φ( x y ∗ ) = 0. These equalities imply the range relationsran(Φ( x y ∗ )) = ran(Φ( x y ∗ )) ⊥ ran(Φ( x y ∗ )) , (28) Accepted in
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Quantum2018-01-16, click title to verify hich imply Φ( x y ∗ ) ∗ Φ( x y ∗ ) = 0. It similarly follows that Φ( x y ∗ )Φ( x y ∗ ) ∗ = 0. Claim 3.
For unit vectors x , x , y , y ∈ C n with h x , x i = h y , y i = 0, it holds thatΦ( x y ∗ ) ∗ Φ( x y ∗ ) = Φ( x y ∗ ) ∗ Φ( x y ∗ ), and Φ( x y ∗ )Φ( x y ∗ ) ∗ = Φ( x y ∗ )Φ( x y ∗ ) ∗ . (29)For the first equality, x + x ⊥ x − x , so Claim 2 implies0 = Φ(( x + x ) y ∗ ) ∗ Φ(( x − x ) y ∗ ) (30)= Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) + Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) (31)= Φ( x y ∗ ) ∗ Φ( x y ∗ ) − Φ( x y ∗ ) ∗ Φ( x y ∗ ) , (32)where we have used Claim 2 to determine that Φ( x y ∗ ) ∗ Φ( x y ∗ ) = Φ( x y ∗ ) ∗ Φ( x y ∗ ) = 0.It may be similarly reasoned that Φ( x y ∗ )Φ( x y ∗ ) ∗ = Φ( x y ∗ )Φ( x y ∗ ) ∗ .We now use Claims 1 to 3 to construct the explicit structure of Φ by examining howit acts on elementary matrices. By Claim 1 it holds thatΦ( E , ) ∗ Φ( E , ) = Φ( E a, ) ∗ Φ( E a, ) and Φ( E , )Φ( E , ) ∗ = Φ( E ,b )Φ( E ,b ) ∗ (33)for all 1 ≤ a, b ≤ n , and hence there exists partial isometries U a : ran(Φ( E , )) → ran(Φ( E a, )), and V b : ran(Φ( E , ) ∗ ) → ran(Φ( E ,b ) ∗ ) (34)for which Φ( E a, ) = U a Φ( E , ) and Φ( E ,b ) = Φ( E , ) V ∗ b (where U and V may be takento be the orthogonal projections onto ran(Φ( E , )) and ran(Φ( E , ) ∗ ) respectively). Noteas well that, statement 5 gives that Φ( E a, ) ∗ Φ( E a , ) = Φ( E ,b )Φ( E ,b ) ∗ = 0 for a = a and b = b , and hence the sets of partial isometries { U a } na =1 , { V b } nb =1 have mutually orthogonalranges.Next, we claim that for all 1 ≤ a, b ≤ n it holds that Φ( E a,b ) = U a Φ( E , ) V ∗ b . In theprevious claim this fact is established when at least one of a and b is 1, so we may assumethat both a, b ≥
2. In this case, Claim 3 implies thatΦ( E , ) ∗ Φ( E ,b ) = Φ( E a, ) ∗ Φ( E a,b ) , (35)and so Φ( E , ) ∗ Φ( E , ) V ∗ b = Φ( E , ) ∗ U ∗ a Φ( E a,b ) . (36)As ran( U ∗ a ) = ran(Φ( E , )) we may cancel the Φ( E , ) ∗ from the left-side of the above equa-tion to get that Φ( E , ) V ∗ b = U ∗ a Φ( E a,b ) (alternatively, we may multiply on the left by thepseudo-inverse of Φ( E , ) ∗ ). Finally, we have ran( U a ) = ran( E a,b ), as Φ( E a,b ) ∗ Φ( E a,b ) =Φ( E a, ) ∗ Φ( E a, ), and so Φ( E a,b ) = U a U ∗ a Φ( E a,b ) = U a Φ( E , ) V ∗ b , (37)as required.The last step is to show that the structure we have just deduced for Φ is the same asthat in statement 7. Let Φ( E , ) = P ri =1 s i x i y ∗ i be a singular value decomposition. Define σ = P ri =1 s i E i,i ∈ M r , which is clearly positive, and define matrices U, V : C n ⊗ C r → C m to act as U ( e a ⊗ e i ) = U a x i , and V ( e b ⊗ e j ) = V b y j . (38)We may verify that these are in fact isometries: h U ( e b ⊗ e j ) , U ( e a ⊗ e i ) i = h U b x j , U a x i i = δ a,b h x j , x i i = δ a,b δ i,j , (39) Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify here we have used that U a and U b have orthogonal ranges for a = b . Hence, U is anisometry as it sends an orthonormal basis to an orthonormal set. The same proof showsthat V is an isometry. Finally, we have thatΦ( E a,b ) = U a Φ( E , ) V ∗ b = r X i =1 s i U a x i y ∗ i V ∗ b = r X i =1 s i U ( E a,b ⊗ E i,i ) V ∗ = U ( E a,b ⊗ σ ) V ∗ . (40)Hence, Φ has the desired form, and the last thing we need is that Tr( σ ) = 1. The finalassumption is that there exists X ∈ M n \ { } with k Φ( X ) k = k X k . This gives k X k = k Φ( X ) k = k U ( X ⊗ σ ) V ∗ k = k X k Tr( σ ) , (41)and hence Tr( σ ) = 1 as desired. Remark 6.
Consider an additional special case of Theorem 3 when Φ is Hermiticitypreserving. As in the proof of the case when Φ is positive, there exists a positive integer r ≤ m/n , a Hermitian H ∈ M r , and an isometry U : C n ⊗ C r → C m for which Φ( X ) = U ( X ⊗ H ) U ∗ for all X ∈ M n . If m < n , then necessarily r = 1 and hence H = ±
1. Itfollows that either Φ or − Φ is a reversible quantum channel. If m ≥ n , then by consideringthe Hahn decomposition of H , one may verify that this form is equivalent to the statementthat there exists reversible quantum channels Φ , Φ : M n → M m with orthogonal rangesand a number r ∈ [0 ,
1] for which Φ = r Φ − (1 − r )Φ . (42) We now prove the main result. As mentioned in the introduction, the inequality in Equa-tion (43) below is known in more generality in C ∗ -algebras [3, Exercise 3.10]. Theorem 7.
Let
Φ : M n → M m be linear with k Φ k = 1 . It holds that k Φ ⊗ id k k ≤ k, (43) with equality if and only if n, m ≥ k , and for any pair of unit vectors u, v ∈ C n ⊗ C k satisfying k (Φ ⊗ id k )( uv ∗ ) k = k (44) (of which at least one such pair must exist), the following statements hold:1. u and v are maximally entangled; i.e. they decompose as u = r k k X a =1 u a ⊗ e a and v = r k k X b =1 v b ⊗ e b (45) for orthonormal sets { u a } ka =1 , { v b } kb =1 ⊂ C n .2. Defining isometries U, V : C k → C n as U = k X a =1 u a e ∗ a and V = k X b =1 v b e ∗ b , (46) there exists a complete trace-norm isometry Ψ : T n ( U M k V ∗ ) → M m , where T n ( U M k V ∗ ) = { ( U XV ∗ ) T : X ∈ M k } , (47) for which Φ( X ) = Ψ( X T ) for all X ∈ U M k V ∗ . Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify roof. Letting u, v ∈ C n ⊗ C k be unit vectors, we will first show that k (Φ ⊗ id k )( uv ∗ ) k ≤ k, (48)which will prove Equation (43). We may assume without loss of generality that u and v have decompositions of the form u = r X a =1 α a u a ⊗ e a , and v = r X b =1 β b v b ⊗ e b , (49)for r ≤ min( k, n ), unit vectors α, β ∈ C r with non-negative entries, and orthonormal sets { u a } ra =1 , { v b } rb =1 ⊂ C n . We have k (Φ ⊗ id k )( uv ∗ ) k = (cid:13)(cid:13)(cid:13)(cid:13) r X a,b =1 α a β b Φ( u a v ∗ b ) ⊗ E a,b (cid:13)(cid:13)(cid:13)(cid:13) (50) ≤ r X a,b =1 α a β b k Φ( u a v ∗ b ) k (51) ≤ r X a,b =1 α a β b k u a v ∗ b k (52)= r X a,b =1 α a β b (53)= h r , α ih r , β i (54) ≤ k r k k α kk β k (55)= r (56) ≤ k (57)where 1 r ∈ C r is the vector of all ones. Hence, it holds that k Φ ⊗ id k k ≤ k .We now examine equality conditions. Suppose that k (Φ ⊗ id k )( uv ∗ ) k = k for unitvectors u, v ∈ C n ⊗ C k with decompositions as in Equation (49). First, we may concludethat r = k , and hence k ≤ n . Furthermore, equality in the application of Cauchy-Schwarzin Equation (55) implies that α = β = q k k , and so u and v are maximally entangled.Thus, defining the isometries U, V : C k → C n as in Equation (46), it holds that (cid:13)(cid:13)(cid:13)(cid:13) k X a,b =1 Φ( U E a,b V ∗ ) ⊗ E a,b (cid:13)(cid:13)(cid:13)(cid:13) = k , (58)and this is equivalent to the more general fact that (cid:13)(cid:13)(cid:13)(cid:13) k X a,b =1 Φ( x a y ∗ b ) ⊗ E a,b (cid:13)(cid:13)(cid:13)(cid:13) = k , (59)for any orthonormal bases { x a } ka =1 ⊂ ran( U ) and { y b } kb =1 ⊂ ran( V ). As k Φ k = 1, theabove implies that k Φ( x a y ∗ b ) k = 1 for all 1 ≤ a, b ≤ k . By looking at 2 × x , x ∈ ran( U ) and y , y ∈ ran( V )with h x , x i = h y , y i = 0, it holds that (cid:13)(cid:13)(cid:13)(cid:13) Φ( x y ∗ ) Φ( x y ∗ )Φ( x y ∗ ) Φ( x y ∗ ) !(cid:13)(cid:13)(cid:13)(cid:13) = 4 . (60) Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify s each block has trace-norm 1, the above 2 × x y ∗ ) ∗ Φ( x y ∗ ) = Φ( x y ∗ )Φ( x y ∗ ) ∗ = 0 (61)for any x , x ∈ ran( U ) and y , y ∈ ran( V ) with h x , x i = h y , y i = 0.This may be written in a more suggestive way: for A, B ∈ U M k V ∗ rank-1, the followingimplications hold:(i) A ∗ B = 0 and A ∗ A = B ∗ B = ⇒ Φ( A )Φ( B ) ∗ = 0, and(ii) AB ∗ = 0 and AA ∗ = BB ∗ = ⇒ Φ( A ) ∗ Φ( B ) = 0.These implications are very similar to statement 5 in Theorem 3, but the adjoints appear indifferent locations. We may remedy this by defining Ψ : T n ( U M k V ∗ ) → M m as Ψ = Φ T n ,where T n is the transpose. We claim that, for A, B ∈ T n ( U M k V ∗ ) rank-1, the followingimplications hold:(a) A ∗ B = 0 and A ∗ A = B ∗ B = ⇒ Ψ( A ) ∗ Ψ( B ) = 0, and(b) AB ∗ = 0 and AA ∗ = BB ∗ = ⇒ Ψ( A )Ψ( B ) ∗ = 0.We will prove that (ii) ⇒ (a), with (i) ⇒ (b) being similar. Let A T , B T ∈ T n ( U M k V ∗ ) berank-1. The statements ( A T ) ∗ B T = 0 and ( A T ) ∗ A T = ( B T ) ∗ B T are equivalent to AB ∗ = 0and AA ∗ = BB ∗ , so (ii) implies that Φ( A ) ∗ Φ( B ) = 0, which is in turn equivalent toΨ( A T ) ∗ Ψ( B T ) = 0.Thus, by Theorem 3, Ψ = Φ T n is a complete trace-norm isometry on T n ( U M k V ∗ ), asrequired. As a corollary to Theorems 3 and 7, we provide two characterizations of the set oflinear maps
Φ : M n → M m satisfying ||| Φ ||| = n k Φ k . Corollary 8.
Let
Φ : M n → M m be linear with m ≥ n . The following are equivalent.1. k Φ k = 1 and ||| Φ ||| = n .2. k J (Φ) k = n and k J (Φ T n ) k = n .3. There exists a complete trace-norm isometry Ψ : M n → M m for which Φ = Ψ T n .In the above, if Φ is Hermiticity preserving so is Ψ , and if Φ is positive then so is Ψ (andhence is a reversible quantum channel).Proof. It is immediate that 3 ⇒ ⇒
2. That 1 ⇒ ⇒ T n , if Φ is Hermiticitypreserving so is Ψ, as it is a composition of Hermiticity preserving maps. The same logicapplies if Φ is positive; with Ψ being a reversible quantum channel following from thepositive case of Theorem 3. Note that Theorem 3 as stated only applies to maps whose domain is all of M n . Here, the domain ofΨ is T n ( UM k V ∗ ) = V M k U T ⊂ M n , so technically we are applying Theorem 3 to conclude that the linearmap X Ψ( V T XU ) is a complete trace-norm isometry on M k . However, this is equivalent to Ψ being acomplete trace-norm isometry on T n ( UM k V ∗ ). Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify A uniqueness result for the Werner-Holevo channels in single-shotquantum channel discrimination
A fundamental operational task in quantum information is to determine which quantumchannel, from a set of possible channels, is acting on a system. The simplest version ofthis task is single-shot quantum channel discrimination , where the goal is to determinewhich of two channels is acting on a system given only a single use. Various aspects ofthis task have been extensively studied (see e.g. [11, 18–25]). For completeness, we give adescription of the task below. Proofs of all facts summarized may be found in [9, Chapter3]. Single-shot quantum channel discrimination may be formulated as a single-player gamespecified by a triple ( λ, Γ , Γ ) , where λ ∈ [0 , and Γ , Γ : M n → M m are quantumchannels. In the game, the player knows a description of the triple ( λ, Γ , Γ ) , and thegame proceeds as follows:1. The referee samples a bit α ∈ { , } with probability p (0) = λ , p (1) = 1 − λ .2. The player is given a single use of the channel Γ α ; i.e. the player gives a quantumstate on the input system of their choice to the referee, who then returns the outputof Γ α .3. The player must guess α (after say, making a measurement on the output).The goal of the player is to maximize the probability that they guess α correctly.Ultimately, all the player can do is prepare an input state to Γ α then try to discriminatethe two possible outputs. In the most unconstrained version of the game, the player is freeto use an auxilliary system; i.e. they can prepare a bipartite quantum state ρ ∈ M n ⊗ M k then discriminate the outputs (Γ α ⊗ id k )( ρ ) . Due to the Holevo-Helstrom theorem forsingle-shot quantum state discrimination [26, 27], the optimal probability of success givena choice of state ρ ∈ M n ⊗ M k is
12 + 12 (cid:13)(cid:13) λ (cid:0) Γ ⊗ id k (cid:1) ( ρ ) − (1 − λ ) (cid:0) Γ ⊗ id k (cid:1) ( ρ ) (cid:13)(cid:13) . (62) Thus, for a fixed auxiliary system of dimension k , the optimal success probability of win-ning the game is given by the optimization of the above expression over density matrices,which reduces to
12 + 12 k λ Γ ⊗ id k − (1 − λ )Γ ⊗ id k k ,H , (63) where, for Ψ : M n → M m , k Ψ k ,H = max {k Ψ( H ) k : H ∈ M n , k H k = 1 , H = H ∗ } . (64) Hence, the optimal value over unconstrained strategies, which amounts to optimizingEquation (63) over k ≥ , is given by
12 + 12 ||| λ Γ − (1 − λ )Γ ||| . (65) A natural question to ask in this setting is: What channel discrimination games (withsystem input size n ) have the maximum possible gap between the optimal performancewith and without entanglement? By Equations (63) and (65) , this amounts to character-izing the games ( λ, Γ , Γ ) , where Γ , Γ : M n → M m , with maximal gap between thenorms k λ Γ − (1 − λ )Γ k ,H and ||| λ Γ − (1 − λ )Γ ||| . (66) Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify n this section we make partial progress towards answering this question. We apply theresults of the previous section to characterize the games (with input space M n ) thathave maximal gap between the norms k λ Γ − (1 − λ )Γ k and ||| λ Γ − (1 − λ )Γ ||| . As itgenerically holds that ≤ k λ Γ − (1 − λ )Γ k ≤ ||| λ Γ − (1 − λ )Γ ||| ≤ , (67) and ||| λ Γ − (1 − λ )Γ ||| ≤ n k λ Γ − (1 − λ )Γ k , the maximum possible gap occurs when ||| λ Γ − (1 − λ )Γ ||| = n k λ Γ − (1 − λ )Γ k . (68) Operationally, the above relations say that the game can be won with certainty usingarbitrary entanglement, but is hard to win without entanglement, with the upper boundon unentangled performance given by k λ Γ − (1 − λ )Γ k being as small as possible giventhat the game can be won with certainty.As detailed in [9, Example 3.36], the Werner-Holevo channels [28] provide a well-knownfamily of channel discrimination games satisfying Equation (68) . For n ≥ , we denotethem as Φ (0) n , Φ (1) n : M n → M n , and they are defined to act as Φ (0) n ( X ) = 1 n + 1 (Tr( X ) n + X T ), and Φ (1) n ( X ) = 1 n − X ) n − X T ) (69) for all X ∈ M n . For the probability λ n = n +12 n , it holds that λ n Φ (0) n − (1 − λ n )Φ (1) n = 1 n T n , (70) and as such (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λ n Φ (0) n − (1 − λ n )Φ (1) n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = n (cid:13)(cid:13)(cid:13) λ n Φ (0) n − (1 − λ n )Φ (1) n (cid:13)(cid:13)(cid:13) . (71) Thus, for n ≥ the triple (cid:0) λ n , Φ (0) n , Φ (1) n (cid:1) is an example of a channel discrimination gamesatisfying Equation (68) . We now apply the result of the previous section to show thatthe game specified by the triple (cid:0) λ n , Φ (0) n , Φ (1) n (cid:1) is in some sense the unique game (withchannels having domain M n ) satisfying Equation (68) . Theorem 9.
Let m ≥ n , Γ , Γ : M n → M m be quantum channels, λ ∈ (0 , be aprobability, and let Φ (0) n , Φ (1) n : M n → M n be the Werner-Holevo channels as given inEquation (69) . It holds that ||| λ Γ − (1 − λ )Γ ||| = n k λ Γ − (1 − λ )Γ k (72) if and only if: • For m < n , there exists a reversible quantum channel Ψ : M n → M m for whicheither ( λ, Γ , Γ ) = ( λ n , ΨΦ (0) n , ΨΦ (1) n ) , or ( λ, Γ , Γ ) = (1 − λ n , ΨΦ (1) n , ΨΦ (0) n ) . • For m ≥ n , there exists r ∈ [0 , and two reversible channels Ψ , Ψ : M n → M m with orthogonal ranges for which λ = rλ n + (1 − r )(1 − λ n ) , and λ Γ = rλ n Ψ Φ (0) n + (1 − r )(1 − λ n )Ψ Φ (1) n , (73) and (1 − λ )Γ = r (1 − λ n )Ψ Φ (1) n + (1 − r ) λ n Ψ Φ (0) n . (74) While the Werner-Holevo channels were originally introduced in [28] for other reasons, in the settingof quantum channel discrimination, they may be intuitively viewed as a way of smuggling the transposeinto the problem, with the properties of the resulting game being inherited from norm properties of thetranspose.
Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify emark 10. Theorem 9 may be interpreted as saying that the game ( λ n , Φ (0) n , Φ (1) n )uniquely satisfies Equation (72) in the following sense: Any game ( λ, Γ , Γ ), whose chan-nels have domain M n and satisfy Equation (72), is constructed out of, and is reducible bythe player to, the game ( λ n , Φ (0) n , Φ (1) n ) in a way that perfectly preserves success probabil-ities. Indeed, mathematically, one can check that k λ (Γ ⊗ id k )( X ) − (1 − λ )(Γ ⊗ id k )( X ) k (75)= (cid:13)(cid:13) λ n (cid:0) Φ (0) n ⊗ id k (cid:1) ( X ) − (1 − λ n ) (cid:0) Φ (1) n ⊗ id k (cid:1) ( X ) (cid:13)(cid:13) (76)for all integers k ≥ X ∈ M n ⊗ M k . Operationally, the construction/reductionof such games ( λ, Γ , Γ ) in terms of ( λ n , Φ (0) n , Φ (1) n ) goes as follows. • For the case m < n , the construction and reduction are natural; up to a reversiblequantum channel (which the player can undo) and a relabeling of the channels (whichthe player knows), the game ( λ, Γ , Γ ) is exactly the game (cid:0) λ n , Φ (0) n , Φ (1) n (cid:1) . • For the case m ≥ n , the relation between ( λ, Γ , Γ ) and (cid:0) λ n , Φ (0) n , Φ (1) n (cid:1) is lessclear, though it can be thought of as a convex combination of relabelings of thegame (cid:0) λ n , Φ (0) n , Φ (1) n (cid:1) , where the player is able to detect which labeling is beingused. Specifically, with probability r , Γ acts as Φ (0) n and Γ acts as Φ (1) n , and withprobability (1 − r ) the labels are reversed. As Ψ and Ψ have orthogonal ranges, theplayer is able to measure which labelling is being used without disturbance. Oncethis is done, the situation from the players perspective is now the same as in thecase m < n , and they may act accordingly. Before proving Theorem 9, we prove a lemma regarding the uniqueness of certain de-compositions of Hermiticity preserving maps into differences of completely positive maps.
Lemma 11.
Let
Φ : M n → M m be Hermiticity preserving, Ψ , Ψ : M n → M m becompletely positive and satisfy Φ = Ψ − Ψ and ||| Φ ||| = ||| Ψ ||| + ||| Ψ ||| , (77) and let u ∈ C n ⊗ C n be a unit vector satisfying ||| Φ ||| = k (Φ ⊗ id n )( uu ∗ ) k . It follows that ||| Ψ ||| = k (Ψ ⊗ id n )( uu ∗ ) k and ||| Ψ ||| = (cid:13)(cid:13) (Ψ ⊗ id n )( uu ∗ ) (cid:13)(cid:13) , (78) and for any other completely positive maps Ψ , Ψ : M n → M m satisfying the conditionsin Equation (77) , (Ψ ⊗ id n )( uu ∗ ) = (Ψ ⊗ id n )( uu ∗ ) and (Ψ ⊗ id n )( uu ∗ ) = (Ψ ⊗ id n )( uu ∗ ) . (79) Hence, if such a u exists with full Schmidt-rank, the completely positive maps Ψ , Ψ satisfying Equation (77) are unique (if they exist).Proof. We have ||| Φ ||| = k (Φ ⊗ id n )( uu ∗ ) k (80)= k (Ψ ⊗ id n )( uu ∗ ) − (Ψ ⊗ id n )( uu ∗ ) k (81) ≤ k (Ψ ⊗ id n )( uu ∗ ) k + k (Ψ ⊗ id n )( uu ∗ ) k (82) ≤ ||| Ψ ||| + ||| Ψ ||| (83)= ||| Φ ||| . (84) Accepted in
Quantum2018-01-16, click title to verify
Quantum2018-01-16, click title to verify ence, all inequalities are equalities, and therefore ||| Ψ ||| = k (Ψ ⊗ id n )( uu ∗ ) k and ||| Ψ ||| = k (Ψ ⊗ id n )( uu ∗ ) k .Next, as Ψ and Ψ are completely positive, it holds that (Ψ ⊗ id n )( uu ∗ ) ≥ ⊗ id n )( uu ∗ ) ≥
0, and so equality in Equation (82) implies that(Φ ⊗ id n )( uu ∗ ) = (Ψ ⊗ id n )( uu ∗ ) − (Ψ ⊗ id n )( uu ∗ ) (85)is the Hahn decomposition of (Φ ⊗ id n )( uu ∗ ). Thus, for any other completely positivemaps Ψ , Ψ : M n → M m satisfying the hypotheses,(Φ ⊗ id n )( uu ∗ ) = (Ψ ⊗ id n )( uu ∗ ) − (Ψ ⊗ id n )( uu ∗ ) (86)is also the Hahn decomposition of (Φ ⊗ id n )( uu ∗ ). Equation (79) therefore follows by theuniqueness of the Hahn decomposition.Finally, if u ∈ C n ⊗ C n has full Schmidt-rank, then a linear map Γ : M n → M m is uniquely specified by the matrix (Γ ⊗ id n )( uu ∗ ), and so Equation (79) implies theuniqueness of the pair Ψ and Ψ (assuming such a pair exists). Proof of Theorem 9.
In both cases the “if” part is a matter of verifying Equation (76),where the case m ≥ n requires use of the fact that Ψ and Ψ have orthogonal ranges.Thus, assume we have a channel discrimination triple ( λ, Γ , Γ ) satisfying Equation(72). By Corollary 8, the norm relation implies λ Γ − (1 − λ )Γ = 1 n Ψ T n , (87)for Ψ : M n → M m a Hermiticity preserving complete trace-norm isometry. Remark 6 givesthe following structure for Ψ: • If m < n , either Ψ or − Ψ is a reversible quantum channel. • If m ≥ n , there exists r ∈ [0 ,
1] and Ψ , Ψ : M n → M m reversible quantumchannels with orthogonal ranges for which Ψ = r Ψ − (1 − r )Ψ .In what follows we will work with the form of Ψ in the case m ≥ n , as the case m < n can be subsumed by the case r = 0 or r = 1 when m ≥ n , even though it is not possiblefor two reversible channels Ψ , Ψ : M n → M m to have orthogonal ranges when m < n .Observe the following facts: • n Ψ T n is Hermiticity preserving and decomposes as a difference of CP maps as givenin Equation (87), • (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n Ψ T n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 = ||| λ Γ ||| + ||| (1 − λ )Γ ||| , and • (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n Ψ T n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:13)(cid:13) n (Ψ T n ⊗ id n )( τ n ) (cid:13)(cid:13) , where τ n = n P na,b =1 E a,b ⊗ E a,b ∈ M n ⊗ M n isthe canonical maximally entangled state.When taken together these facts imply, by Lemma 11, that Equation (87) is the unique decomposition of n Ψ T n into a difference of CP maps with the above properties. In theremainder of the proof, we will exhibit a (seemingly) different decomposition of n Ψ T n ,verify that it also satisfies the assumptions of Lemma 11, then conclude that the twodecompositions are necessarily the same. The Hahn decomposition of a Hermitian matrix H ∈ M n is the unique decomposition of H as adifference H = P − Q with P, Q ≥ P Q = 0. For H Hermitian and
P, Q ≥
0, it holds that H = P − Q is the Hahn decomposition of H if and only if k H k = k P k + k Q k . Accepted in
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Quantum2018-01-16, click title to verify ote that n T n = λ n Φ ( n )0 − (1 − λ n )Φ ( n )1 , and hence1 n Ψ T n = ( r Ψ − (1 − r )Ψ )( λ n Φ ( n )0 − (1 − λ n )Φ ( n )1 ) (88)= (cid:2) rλ n Ψ Φ ( n )0 + (1 − r )(1 − λ n )Ψ Φ ( n )1 (cid:3) − (cid:2) (1 − r ) λ n Ψ Φ ( n )0 + r (1 − λ n )Ψ Φ ( n )1 (cid:3) . (89)The maps in the square brackets are completely positive, and satisfy (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) rλ n Ψ Φ ( n )0 + (1 − r )(1 − λ n )Ψ Φ ( n )1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 − r ) λ n Ψ Φ ( n )0 + r (1 − λ n )Ψ Φ ( n )1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (90)= rλ n + (1 − r )(1 − λ n ) + (1 − r ) λ n + r (1 − λ n ) = 1 . (91)Hence, by the uniqueness clause of Lemma 11, Equations (73) and (74) hold.When m < n , either r = 0 or r = 1, in which case either( λ, Γ , Γ ) = (cid:0) λ n , Ψ Φ (0) n , Ψ Φ (1) n (cid:1) , or ( λ, Γ , Γ ) = (cid:0) − λ n , Ψ Φ (1) n , Ψ Φ (0) n (cid:1) , (92)as required. The canonical example of a linear map
Φ : M n → M m satisfying k Φ k = 1 and k Φ ⊗ id k k = k (93) is the matrix transpose, and we have proven that, up to an equivalence, the transpose isthe unique map satisfying the above equation. We have applied this result in the setting ofsingle-shot quantum channel discrimination to prove that a channel discrimination game ( λ, Γ , Γ ) (with input dimension n ) satisfies the norm relation ||| λ Γ − (1 − λ )Γ ||| = n k λ Γ − (1 − λ )Γ k (94) if and only if it is in some sense equivalent to the game ( λ n , Φ (0) n , Φ (1) n ) , where Φ (0) n , Φ (1) n are the Werner-Holevo channels, and λ n = n +12 n .The uniqueness result for the Werner-Holevo channel discrimination game is almost,but not quite, a characterization of channel discrimination games with maximal gap be-tween the optimal performance of entangled and unentangled strategies. Characterizingsuch games requires an understanding of the maximal gap between k Φ k ,H and ||| Φ ||| for Hermiticity preserving linear maps Φ : M n → M m . For example, is it true that ||| Φ ||| ≤ n k Φ k ,H ? More generally, is it true that k Φ ⊗ id k k ,H ≤ k k Φ k ,H ? (95) It is not so clear if the proof of the inequality in Theorem 7 can be adapted to this situation.It seems reasonable to conjecture that some inequality of the above form holds, and thatthe transpose will uniquely saturate the inequality.Another natural question is whether the characterization of linear maps
Φ : M n → M m for which k Φ k = 1 and k Φ ⊗ id k k = k holds approximately. E.g. if k Φ k = 1 and ||| Φ ||| ≥ n − (cid:15) , then is Φ necessarily close in some sense to the transpose (followed by acomplete trace-norm isometry)? Accepted in
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Quantum2018-01-16, click title to verify cknowledgements We thank John Watrous, Vern Paulsen, and Chi-Kwong Li for helpful discussions andcomments. This work was supported by NSERC, the Ontario Graduate Scholarship, andthe Queen Elizabeth II Graduate Scholarship in Science and Technology.
A Duality of the operator norm and trace-norm
Let k · k denote the operator norm for matrices, as well as the induced operator norm forlinear maps of matrices, i.e. for linear
Φ : M n → M m , k Φ k = max {k Φ( X ) k : X ∈ M n , k X k = 1 } . (96) For
A, B ∈ M n , we denote the Hilbert-Schmidt inner product as h A, B i = Tr( A ∗ B ) , andfor a linear map Φ : M n → M m , we use Φ ∗ to denote the adjoint of Φ with respect to thisinner-product. That is, Φ ∗ : M m → M n is the unique linear map satisfying h A, Φ( B ) i = h Φ ∗ ( A ) , B i (97) for all A ∈ M m and B ∈ M n .For our purposes, the “duality” of the trace-norm and operator norm may be summa-rized by the following: For any matrix A ∈ M n , it holds that k A k = max {|h X, A i| : X ∈ M n , k X k ≤ } , (98) and k A k = max {|h X, A i| : X ∈ M n , k X k ≤ } . (99) A direct implication of these expressions is that, for any linear
Φ : M n → M m , it holdsthat k Φ k = k Φ ∗ k and k Φ k = k Φ ∗ k . (100) The above relations enable interconversion of facts about the trace-norm and factsabout the operator norm. For example, the statement k Φ ⊗ id k k ≤ k k Φ k for all linear maps Φ : M n → M m (101) is equivalent to the statement k Φ ⊗ id k k ≤ k k Φ k for all linear Φ : M n → M m . (102) This is why [3, Exercise 3.10], which directly generalizes the statement in Equation (102) to arbitrary linear maps between unital C ∗ -algebras, is referenced in the introduction as ageneralization of the statement in Equation (101) . Similarly, our main result characterizingthe linear maps Φ : M n → M m for which k Φ ⊗ id k k = k k Φ k , may also be translated intoa characterization of such maps for which k Φ ⊗ id k k = k k Φ k . References [1] R. Smith. Completely Bounded Maps between C ∗ -Algebras. Journal of the LondonMathematical Society , s2-27(1):157–166, 1983. DOI: 10.1112/jlms/s2-27.1.157.[2] J. Tomiyama. Recent Development of the Theory of Completely Bounded Maps be-tween C ∗ -Algebras. Publications of the Research Institute for Mathematical Sciences ,19(3):1283–1303, 1983. DOI: 10.2977/prims/1195182030.
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