ENTANGLEMENT BREAKING QUANTUM CHANNELS AND THE PPT-SQUARED CONJECTURE

Transcription

1 ENTANGLEMENT BREAKING QUANTUM CHANNELS AND THE PPT-SQUARED CONJECTURE NICHOLAS A. MANOR Abstract. Entanglement is a valuable resource in quantum information, therefore we would like to better understand operations on states which completely destroy entanglement. In practice, a certain family of quantum channels called positive partial transpose (PPT) maps are easier to implement as they require less energy. Since quantum states may pass through many channels before reaching their final destination, there is an important question we must ask which was first posed by M. Christandl in [1]: do products of PPT channels break entanglement? Experimental data would suggest that this is the case; this is known as the PPT-squared conjecture. Herein we discuss the general theory of entanglement breaking quantum channels, and add a new characterization in terms of factoring through an Abelian C*-algebra. We also answer the asymptotic analogue to the conjecture in the affirmative: do powers of a PPT map converge to the set of entanglement breaking channels? Finally, we examine a one parameter family of maps on matrix algebras defined using Schur multipliers. It is shown that when these maps are PPT then their product is necessarily entanglement breaking, answering the conjecture for at least a large family of operators. 1. Overview of Quantum States and Quantum Channels Of central importance to the study of quantum information is the study of states, and how they are altered by certain quantum processes such as quantum channels and measurements. In the general theory of C*-algebras we define states as being positive linear functionals of norm 1. In the case of a unital C*-algebra, the latter condition may be replaced with the condition that they must send the identity element to 1 (Note: this is not an immediate fact). In quantum information we generally deal with finite dimensional full complex matrix algebras, and we have various equivalent ways to describe states in this setting. By the Riesz representation theorem, using the Hilbert- Schmidt inner product on M n, a linear functional ρ takes the form A tr(ap ) where tr : M n C denotes the normalized trace and P is some fixed matrix. Now we may ask the following question: what matrices P correspond to states ρ on M n? If P is not a positive matrix then there must exist some vector v C n such that P v, v = v P v 0, in which case tr(vv P ) 0. 1

2 2 So clearly, if we desire ρ to be positive we require P to be a positive matrix. Conversely, if P is positive then it has a unique positive square root P and thus we have ρ(a) = tr(ap ) = tr( P A P ) which is clearly a positive functional (since positive matrices have non-negative trace). Note that ρ being unital means precisely that ρ(i) = tr(p ) = 1. So the previous paragraph tells us that positive functionals correspond exactly to positive matrices, and together with the above remark we have that states correspond to positive matrices with trace 1. As a result we define a quantum state in M n to be a positive matrix with trace 1. We now have the full power of linear algebra at our disposal in studying these objects. Note: These matrices are also referred to as density matrices by quantum information theorists, since they may be used along with a given quantum measurement system to produce a finite probability density function. It is important to mention the sub-concept of a pure state, which is exactly an extreme point in the set of states. We know that the set of states is bounded, and it is clearly closed and convex. By the Krein-Milman theorem we then know that all states on M n may be written as a linear combination of pure states, so it is worthwhile to understand these particular maps. States which are not pure are called mixed states. Recall that by the spectral theorem, we may decompose a quantum state P as λ 1 v 1 v1 + + λ nv n vn where the v i are (mutually orthogonal) unit vector, and the λ i are some non-negative numbers. Note that since P is a quantum state we require that λ λ n = 1, which proves that P is a convex combination of states defined by rank one projections. We ll henceforth refer to such states as vector states, and use the notation ρ v to refer to the state A tr(avv ) = v Av. The above discussion would suggests that the set of vector states is the set of extreme points, and this is indeed true. For one inclusion note that every state corresponding to a quantum state which has rank strictly greater than one may be decomposed into a convex combination in the manner illustrated above. This says that all extreme points are necessarily vector states. As for the opposite inclusion, suppose we are given a vector state ρ v and two states ρ 1 and ρ 2 with associated quantum states P 1 and P 2 such that ρ v = 1 2 ρ ρ 2. Then we must have vv = 1 2 (P 1 + P 2 ). Since rank(p 1 + P 2 ) rank(p i ) we necessarily have that the P i have rank one, and since their trace is also one then they are both rank one projections. Let s write them as P i = v i vi and note that if {v 1, v 2 } is linearly dependent then we are done, because P 1 = P 2 in this case. If they form a linearly independent set then rank(vv ) = rank(p 1 + P 2 ) = 2, a contradiction which establishes the following proposition.

3 Proposition 1.1. The extreme points of the set of states on M n are precisely the vector states. Likewise, the extreme points of the set of quantum states in M n are precisely the rank one projection. Now that we have a precise notion of quantum state, we may properly discuss processes involving these objects. We will focus our attention to quantum channels. These maps should model transitions from a state in one lab to a state in possibly another lab, so at the very least they should send positive matrices to positive matrices between full matrix algebras and be trace preserving. We usually denoted quantum channels by the uppercase Greek letters Φ, Ψ or Γ. These two conditions are not quite enough in general, because of the possibility that two labs, A and B, have states which are entangled with one another. Loosely, this means that the induced probability distribution on A given some measurement system cannot be fully described without also understanding B. Mathematically, entanglement is modelled using tensor products. We ll refer to states in a tensor product M n M m as bipartite states. This introduces a new condition: if a quantum channel Φ acts on the state of a lab A which is entangled with B, we would like the resulting matrix to be a valid joint state. For this to always occur we require that Φ id B also be positive and trace preserving, where id B is the identity on the full matrix algebra associated to B. Note that both Φ and id B are trace preserving, and so too is their tensor product. To see this, observe that the trace of a pure tensor matrix P Q is given by 3 tr(p Q) = i,j p ii q jj = tr(p ) tr(q). So Φ id B will be trace preserving on pure tensors, and since everything in sight is linear then they are trace preserving on the entire tensor product. The problem of requiring this tensor product to be positive is more interesting, and will be addressed in greater detail in the coming section Completely Positive Maps. In the theory of C*-algebras (or operator systems more generally) there is a very well developed theory of positivity. The notion of positivity induces an order structure on an operator system, and in this section we ll study maps which preserve this order structure while restricting our attention mostly to maps between full matrix algebras. A map Φ : M n M m is called positive if it sends positive matrices to positive matrices, and we call it k-positive if Φ id k : M n M k M m M k is positive. Finally, it is completely positive if it is k-positive for all k. Definition 1.2. We call a linear map Φ : M n M m a quantum channel if it is completely positive and trace preserving (CPTP).

4 4 The study of CP maps is central to modern operator theory, and so we luckily have many useful tools at our disposal when studying quantum channels. Recall that M n can be given a Hilbert space structure using the Hilbert- Schmidt inner product, i.e., given matrices A and B we define A, B := tr(ab ). A simple computation shows that this makes the standard basis matrices E ij into an orthonormal basis, hence this inner product induces the Euclidean norm on M n viewed as C n2. This process of turning a C*-algebra into an inner product space works more generally for faithful states on a unital C*-algebra. Positivity in M n behaves nicely with respect to the inner product; this, in essence, follows from the trace property that tr(ab) = tr(ba). Let s formalize this observation in two lemmas. Lemma 1.3. A matrix P M n is positive if and only if P, Q 0 for every positive Q M n. Proof. Suppose first that P 0 and let Q be any positive matrix. Then it has a unique positive square root Q, so P, Q = tr( QP Q) 0 since QP Q 0. For the other direction, recall that P is positive if and only if P v, v 0 for all v C n. Assume that P is not positive, then there exists a v so that P v, v 0. From this it follows that P, vv = tr(p vv ) = tr(v P v) = v P v = P v, v 0, which is a contradiction since vv 0. We also have that, for positive matrices, orthogonality with respect to the Hilbert space structure is compatible with orthogonality of ranges. This again follows from the trace property. Lemma 1.4. Let P, Q M n be positive matrices. Then P Q = 0 if and only if P, Q = 0. Proof. The forward direction is trivial: supposing P Q = 0, we have P, Q = tr(p Q) = tr(0) = 0.

5 Now suppose P, Q = 0. Recall that P and Q have unique positive square roots P and Q. With these we perform the following computation: P, Q = tr(p Q) = tr(( P Q) ( P Q)). But by hypothesis this is zero. So since the trace is faithful we must have ( P Q) ( P Q) = 0, and hence P Q = 0. Multiplying by the square roots again on either side yields P Q = 0. The following lemma and its corollary are simple and very useful facts that are worth noting, as they ll come in handy it some proofs later on. Lemma 1.5. If Θ : M n M m is a positive map, then so is Θ where the adjoint is taken with respect to the Hilbert-Schmidt inner products on M n and on M m. Proof. Let P M n and Q M m be positive. Then Θ (Q), P = Q, Θ(P ) 0 since it is an inner product of two positive matrices. By Lemma 1.3 we know Θ is positive. Corollary 1.6. If Θ : M n M m is k-positive, then so is Θ. Proof. By the lemma above we know that (Θ id k ) = Θ id k is positive, and the result follows. It seems strange at first that positivity should break down when applying a positive map to blocks of a larger matrix, but there are simple examples which do exactly this. For instance, we can easily show that the transpose map T : M 2 M 2 is positive but not 2-positive (hence not completely positive either). We know that 2 i,j=1 E ij E ij is a positive scalar multiple of a projection, so it is positive. Applying T id 2 to it produces 2 i,j=1 E ji E ij which has e 1 e 2 e 2 e 1 as an eigenvector with eigenvalue -1. Choi s theorem gives us a very easy way, in general, to test whether a given map is completely positive. This statement and proof of the theorem can be found in [9]. Theorem 1.7 (Choi s Theorem). Let Φ : M n M m be a linear map. The following are equivalent: (i) Φ is completely positive. (ii) Φ is n-positive. (iii) The Choi matrix of Φ C Φ := (Φ(E ij )) n i,j=1 M n M m 5

6 6 is positive. (iv) Φ has the form Φ(A) = V i AV i for a finite set {V i } k of m n matrices. Proof. (i) = (ii): This follows from the definition of complete positivity. (ii) = (iii): This is also immediate since (E ij ) n i,j=1 is a positive matrix. Indeed, it equal to np where P is the rank one projection onto n e i e i. (iii) = (iv): Since C Φ is positive in M nm it may be written as k v ivi for some vectors v i C nm. We may write these vectors as n blocks of m 1 vectors: h i 1 v i =. h i n, where the h i j are vectors in Cm. We use these to define the m n matrices V i as follows: V i := (h i 1 h i n). We ll show that the Choi matrix of the map A k V iavi k v ivi ; this is a routine check: is also ( V i E st V i ) n s,t=1 = = = (V i e s (V i e t ) ) n s,t=1 (h i s(h i t) ) n s,t=1 v i vi. Since maps between full matrix algebras are completely determined by their Choi matrix, the result follows. (iv) = (i):

7 Let l 0 and let (A ij ) l i,j=1 M n M l be positive. Then Φ id l ((A ij )) = (Φ(A ij )) = (V s A ij Vs ) = s=1 s=1 V s I l (A ij )V s I l is clearly still positive as a sum of positive matrices. This proves that Φ is completely positive. The operators in (iii) are referred to as the Kraus operators of the map Φ. These behave nicely with respect to the adjoint, as illustrated by this lemma. Lemma 1.8. If Φ : M n M m is a completely positive map with Kraus operators V 1,..., V k, then Φ is completely positive with Kraus operators V1,..., V k. Proof. That Φ is completely positive follows directly from Corollary 1.6. Taking A M m and B M n we see that Φ (A), B = A, Φ(B) = tr(a i V i BV i ) 7 = tr(( i V i AV i )B) = i V i AV i, B. Since A and B are arbitrary, the result follows. This lemma gives us a nice duality between trace preserving and unital completely positive maps. Corollary 1.9. A completely positive map Φ : M n M m is trace preserving if and only if Φ is unital. Proof. If Φ is trace preserving and has Kraus operators V 1,..., V k then for all A M n we have tr(a) = tr(φ(a)) = tr( V i AVi ) = tr( V i V i A).

8 8 Since the trace is faithful then we must have k previous lemma this is just Φ (I). The converse works the same way. V i V i = I. But by the For the remainder of this chapter we ll examine the two families of quantum channels which are relevant to studying the conjecture Entanglement Breaking Channels and Separable States. As mentioned in the overview, entanglement is modelled using tensor products. It is a valuable resource in quantum information, and as such it is important to understand the class of channels which eliminate entanglement altogether. These maps will form the central study of this project. Definition A positive element ρ of M n M l is called separable if there are k 0, and positive matrices P 1,..., P k M n, Q 1,..., Q k M l such that ρ = P i Q i. In the case that ρ is a quantum state, i.e., has trace one, we sometimes say it is unentangled, if it is not separable we call it entangled. In general it is extremely difficult to determine whether a given state is entangled. In fact, the problem of determining separability is known to be NP-hard [2]. There is nonetheless a test which can be applied to answer this question, but in high dimension this can become equally intractable. Theorem Let ρ be a state in M n M m. Then the following are equivalent: (i) ρ is separable. (ii) Θ id m (ρ) is positive for all positive maps Θ on M n. (iii) id n Θ(ρ) is positive for all positive maps Θ on M m. Proof. We will prove only the easy implication (i) = (ii); the implication for (iii) is identical. Let ρ = i P i Q i be algebraically separable (this suffices since the set of positive matrices is closed), and Θ be a positive map on M n. Then Θ id m (ρ) = Θ(P i ) Q i. i Since all the Θ(P i ) and Q i are positive, then the above matrix is positive. The other implications can be found in [3] and require a clever use of the Hahn-Banach theorem and a characterization of positive maps between matrix algebras given in [5]. From this theorem we see that the maximally entangled state E ij E ij M n M n is not separable; applying the partial transpose map yields T id n ( E ij E ij ) = E ji E ij,

9 which has an eigenvalue of -1 corresponding to eigenvectors of the form e i e j e j e i. Definition A completely positive map Φ : M n M m is called entanglement breaking (EB) if for every k and every positive matrix ρ M n M k we have that Φ id k (ρ) is separable. Horodecki, Shor and Ruskai in [4] prove an analogue of Theorem 1.11 for EB maps which is quite useful; it makes use of Choi s theorem to provide many nice ways to construct entanglement breaking maps, including a correspondence between entanglement breaking quantum channels and unentangled bipartite states given by the equivalence of (i) and (iv) in the theorem below. The equivalence of (vii) was first given in [6], and (viii) is new but follows trivially from (vii). This new characterization gives a nice C*-algebraic description of entanglement breaking maps, and suggests a nice way to generalize the theory of entanglement maps to the abstract operator system setting. Theorem Let Φ : M n M m be a quantum channel. The following are equivalent: (i) Φ is entanglement breaking. (ii) Θ Φ is completely positive for all positive maps Θ on M m. (iii) Φ Θ is completely positive for all positive maps Θ on M n. (iv) The Choi matrix C Φ is separable. (v) Φ admits a set of rank one Kraus operators. (vi) Φ has the form Φ(A) = tr(af i )R i for density matrices R i and a POVM {F i }, i.e., a finite set of positive matrices such that F i = I. (vii) Φ factors through a finite-dimensional Abelian C*-algebra, i.e., there are CP maps Ψ : M n l k and Γ : l k M m such that Φ = Γ Ψ. (viii) Φ factors through an Abelian C*-algebra. Proof. We proceed by first proving that (i), (ii), (iv), (v), and (vi) are all equivalent; then we show (vi) = (vii) = (viii) = (iii) = (v). (i) (ii): If k is a positive integer and ρ is any state in M n M k, then Φ id k (ρ) is separable. By Theorem 1.11 we have that Θ id k (Φ id k (ρ)) = (Θ Φ) id k (ρ) is positive for any positive map Θ : M m M m. Since k and ρ were chosen arbitrarily, then Θ Φ is completely positive. For the other direction let k and ρ be as above. Then Θ id k (Φ id k (ρ)) = (Θ Φ) id k (ρ) is positive, meaning Φ id k (ρ) is separable. So Φ is entanglement breaking. 9

10 10 (ii) (iv): Since Θ Φ is CP for all positive Θ, then (Θ Φ) id k ( E ij E ij ) = Θ id k (Φ id k ( E ij E ij )) is positive. Using Theorem 1.11 again, this tells us that the Choi matrix Φ id k ( E ij E ij ) of Φ is separable. To prove the other direction, simply note that (Θ Φ) id k ( E ij E ij ) would be positive by Theorem 1.11, so Θ Φ is entanglement breaking. (iv) (v): Since Φ has a separable Choi matrix, there are positives R 1,..., R k in M m and F 1,..., F k in M n such that (Φ(E ij )) n i,j=1 = n Φ(E ij ) E ij = i,j=1 R l F l. By rescaling we may assume that the R l are all density matrices. Now note that the map Ψ : M n M m : A k l tr(af l)r l has the same Choi matrix as Φ; since these maps are completely determined by their Choi matrices then they must be equal. All we have left to show is that {F l } forms a POVM. But this follows from Φ being trace preserving. The other direction is proved the same way. (vi) (v): Let E 1,..., E k be rank one Kraus operators for Φ. Then for any positive A, Ei AE i is a positive matrix of rank at most one and its range is contained in the (one-dimensional) range of Ei. So it is a positive scalar multiple ρ i (A) of R i, the projection onto the range of Ei. It is easy to verify that l=1 Φ(A) = ρ i (A)R i for all A M n simply by decomposing it into a linear combination of positives. Now, since the ρ i are all positive functionals then there are positive matrices F i such that ρ i (A) = tr(af i ). We also require Φ to be trace preserving, so {F i } must be a POVM. To prove the other direction, note that by spectrally decomposing F i and R i we may assume they have the form v i vi and w i wi, respectively. From

11 11 this reduction we can easily prove (iv): Φ(A) = tr(av i vi )w i wi = = = = tr(vi Av i )w i wi (vi Av i )w i wi w i (vi Av i )wi (w i vi )A(w i vi ). So Φ has Kraus operators w i vi, which are all rank one. Now that we have shown (i) (ii) (v) (vi) (iv), it remains to show the chain of equivalences (vi) = (vii) = (viii) = (iii) = (v). (v) = (vii): Define Ψ : M n l k and Γ : l k M m by Ψ(A) = (tr(af i )) k and Γ((a i ) k ) = a i R i. We clearly have Φ = Γ Ψ, so we must now show that they are completely positive. Firstly, we know tr(af i ) 0 whenever A 0 by the correspondence between positive matrices and positive functionals. This tells us that Ψ is a positive map, but since its codomain is a commutative C*-algebra then it is completely positive ([8], chapter 3). It is also clear that Γ is positive, and a theorem by Stinespring (which can be found in [8]) tells us that positive maps out of an Abelian C*-algebra are completely positive. (vii) = (viii): This is trivial, since l k is an Abelian C*-algebra. (viii) = (iii): Letting Θ be a positive map on M n, and Ψ : M n A and Γ : A M m be the factors of Φ through an Abelian C*-algebra A, we have Φ Θ = Γ (Ψ Θ). But Ψ Θ is completely positive as it is a positive map into an Abelian C*-algebra ([8]). (iii) = (v): If Φ Θ is completely positive for all Θ, then the adjoint Θ Φ is also completely positive. But Θ runs across all positive maps on M n as Θ does, so Φ satisfies (ii). We have proven above that (ii) is

12 12 equivalent to (iv), hence Φ has rank one Kraus operators, so Φ does as well Positive Partial Transpose (PPT) Channels and States. In this section we introduce a type of quantum channel which, in practice, would require relatively low energy to implement and so it is useful to understand its behaviour well. We say a quantum channel Φ : M n M m is positive partial transpose (PPT) when Φ T n is completely positive, where T n : M n M n is the tranpose map. A state ρ in M n M m is PPT if T n id m (ρ) is again a state. Notice that for any Φ : M n M m we have n T n id m (C Φ ) = T n id m ( Φ(E ij ) E ij ) = i,j n T n Φ(E ij ) E ij i,j=1 = (T n Φ(E ij )) n i,j=1 = C Tn Φ. Hence Φ is PPT iff C Φ is a PPT state This gives us a natural correspondence between PPT channels and bipartite states. In this way we may construct PPT channels from PPT states and these aren t so hard to come up with; for instance all separable states are necessarily PPT since transposition is a positive map (see Theorem 1.11). As mentioned in the previous section, entanglement is a useful resource and we would like to preserve it. A natural question to ask is whether PPT maps break entanglement; fortunately this is not true in general ([10]), but experimental evidence suggests that compositions of PPT channels always break entanglement ([1]). This problem is addressed in the following conjecture by M. Christandl in [1]. Conjecture (PPT-square Conjecture). Let Φ : M n M m and Ψ : M m M k be PPT quantum channels, then their composition Ψ Φ is entanglement breaking. Interestingly, it is actually stronger to ask whether the square of any PPT map Φ : M n M n is entanglement breaking; if the channels Φ 1 : M n M m and Φ : M m M p are PPT then so is Φ : M n+m+p M n+m+p, which is defined as follows: Φ A B = Φ 1 (A) 0. C 0 0 Φ 2 (B)

13 Applying this map twice yields Φ 2 (Φ 1 (A)) in the bottom right corner, and 0 elsewhere. So Φ 2 is entanglement breaking if and only if Φ 2 Φ 1 is. Note that Φ is not trace preserving, so it is not a channel. Remark Christandl s conjecture is about channels, and it is not clear whether it is equivalent to asking the same question for PPT maps in general. It is a convenient fact that Φ is PPT if and only if T m Φ is completely positive, in other words it doesn t matter whether we compose or precompose the transpose map. This can be easily proven from the following observation: C Φ Tn = (Φ(E ji ) = (T m Φ(E ij )) T = C T T m Φ. Using Choi s theorem, and the fact that a matrix is positive if and only if its transpose is, the result follows. From a result of Woronowicz [11], it follows that the conjecture holds for channels on M 2. His result states that every positive map on M 2 is the sum of a completely positive map and a completely copositive map (a completely positive map composed with transpose), so of course PPT maps on M 2 are entanglement breaking by Theorem 1.11 and squares of such maps are again entanglement breaking. In a 2017 paper ([7]), Klep et al prove a nice correspondence between positive maps (resp. completely positive) from M n to M m and complex homogeneous polynomials in n + m variables which are positive (resp. sums of Hermitian squares) via the map Φ y Φ(xx )y, where x is a vector in n complex unknowns and y is a vector in m complex unknowns. To be more precise, this map is a bijection between L(M n, M m ) and the set of symmetric multiforms SymC[x, x, y, ȳ] 1,1,1,1 := span C {x i x j y k ȳ l : i, j, k, l = 1, 2}. A consequence of Woronowicz s result is that the positive polynomials in this set are precisely sums of Hermitian squares and polynomials which are Hermitian squares with the x i and x i switched. An easy example of a channel which is not PPT is the identity map M n M n simply because transposition is not completely positive. In a sense PPT maps unentangle states partially. In particular, if a PPT map Φ : M n M m is acting on a bipartite system via Φ id m : M n M m M m M m then it cannot have maximally entangled states in its range, e.g., m i,j=1 E ij E ij, since these are known not to be PPT. A nice consequence of this is that PPT maps cannot be surjective onto a full matrix algebra. 13

14 14 2. An Asymptotic Answer to the PPT-squared Conjecture It turns out that Christandl s intuition holds asymptotically in the following sense: the sequence of iterates of a PPT channel Φ on M n approaches the set of entanglement breaking maps. To prove this we use some very basic theory of Abelian semigroups. Interestingly, we prove this by first examining the case when Φ : M n M n is an idempotent unital PPT map. Definition 2.1. The Choi-Effros product on the range of Φ is given by a b = Φ(ab). Lemma 2.2. Let Φ : M n M n be an idempotent unital PPT map. Then the range of Φ is an Abelian C*-algebra under the Choi-Effros product. Proof. We would like to conclude first that the range is at least a C*-algebra by applying [8, Theorem 15.2], and for this we need the range to be an injective operator system. We will prove this now. Suppose S is an operator system in a C*-algebra A, and that Ψ : S Φ(M n ) is a CP map. Then we can precompose it with Φ to no effect, i.e., Φ Ψ = Ψ. By Arveson s extension theorem there is a completely positive extension Γ : A M n of Ψ viewed as a map into M n, but Φ Γ then clearly extends Φ Ψ = Ψ viewed as a map into Φ(M n ). Next we ll show that Φ(M n ) is Abelian; if it is non-abelian then it has a direct summand isomorphic to M k for some k 2. So by composing with the associated conditional expectation we get an induced PPT map Ψ which is surjective onto M k. In this case, Ψ id k : M n M k M k M k is also surjective. So there is a positive matrix (A ij ) M n M k with Ψ id k ((A ij )) = (E ij ), however the canonical maximally entangled state is known not to be PPT. This tells us that the range is Abelian. Proposition 2.3. Let Φ be an idempotent, unital PPT map on M n. Then Φ is entanglement breaking. Proof. By the previous lemma, Φ factors through a finite dimensional Abelian C*-algebra; so by Theorem 1.13 we have the result. Lemma 2.4. If Φ is a contractive PPT map then there is an idempotent map Ψ in the limit points of (Φ k ) k 1. Proof. Let S denote the closure of {Φ k : k N}. This is compact since Φ is contractive, and it is an Abelian semigroup under composition. Let K = a S as be the intersection of all singly generated ideals. Then we claim K is a minimal ideal in S. First of all, this set is non-empty by the finite intersection property and by the fact that the product of finitely many ideals is contained in their intersection. It is also clearly an ideal, as an intersection of ideals. Minimality follows from the fact that every ideal contains a singly generated ideal, and of course every singly generated ideal contains K.

15 It will follow from minimality that K has a multiplicative identity, thus giving us an idempotent in S (the identity of K). So take k K. Since K is minimal we have k 2 S = K, so there is s K such that (sk)k = sk 2 = k. We claim that sk is the identity in K. Taking any k K, again by minimality there is s S such that s k = k. From this we get that (sk)k = (sk)s k = s (sk)k = s k = k. So we may take Ψ = sk. Note that if Φ is trace preserving then so is the idempotent Ψ since the trace is continuous. The next lemma tells us how Φ may approach the set of entanglement breaking channels, since the existence of Ψ allows us to generate many more entanglement breaking maps in the limit points of (Φ k ) k 1. This remark is formalized in the final theorem of the chapter. Lemma 2.5. If Φ and Ψ are as above then Φ k Φ k Ψ 0. Proof. Since M n is a finite dimensional complex vector space, we may view Φ as being an upper triangular matrix with its modulus 1 eigenvalues appearing first on the diagonal, then followed by its strictly smaller eigenvalues. Of course, Ψ will also be upper triangular as it is a limit of powers of Φ, and since Ψ is a projection then it must be diagonal with 1 s appearing first and then 0 (corresponding, respectively, to Φ s eigenvalues in T and then in the open disk D). From here it is clear that Spec(Φ ran Ψ ) T and Spec(Φ ran(id Ψ) ) D. So since Φ k Φ k Ψ = Φ k (id Ψ) it may be viewed as an upper triangular matrix with only elements of D on the diagonal, thus it converges to zero as k goes to infinity. Theorem 2.6. If Φ is either a unital or a trace preserving PPT map then the sequence of distances d(φ k, EB) goes to 0, where EB denotes the set of entanglement breaking maps on M n. In other words: Φ is asymptotically entanglement breaking. Proof. In the unital case, we know that the idempotent map Ψ from the lemma above is PPT and hence entanglement breaking by Proposition 2.3. So for every n the map Φ k Ψ is entanglement breaking, and by the previous lemma this implies d(φ k, EB) 0. To retrieve the trace preserving case, recall that by Corollary 1.9 trace preserving maps are precisely the adjoints of unital ones, and the set of entanglement breaking maps is also *-symmetric. So, since the adjoint operation is an isometry we get the result. While this is a very nice result mathematically, it doesn t tell us whether any particular PPT channel or UCP map will square to an entanglement breaking one. The following section will tackle the question directly for a small family of operators, and therein we will show the conjecture to be true within that context. 15

16 16 3. Schur Product Maps Defined by Graphs In this section we study one-parameter families of maps defined using the Schur multiplier of the adjacency matrix of a graph. Schur multipliers are a very well understood and nice class of linear maps; as a result we can answer the PPT-squared conjecture in the affirmative in the setting described below. Given two n m matrices A, B, we define their Schur product to be A B := (a ij b ij ). In a sense it s the most naive way to multiply matrices, and note that this product is commutative. Let s denote by S A the map B A B; it is easy to show that S A is completely positive if and only if A is positive. If A is not positive then S A (E) = A where E is the (positive) matrix of all ones, so S A is not positive, let alone completely positive. If A is positive, then the Choi matrix C SA is given by n (A E ij ) n i,j=1 = (a ij E ij ) n i,j=1 = a ij E ij E ij. i,j=1 This matrix, call it Ã, is clearly Hermitian since A is Hermitian, so it remains to show that it has non-negative spectrum. Note that Ã(e s e t ) = 0 whenever s t, and n Ã(e s e s ) = a ij E ij e s E ij e s Compare this to = i,j=1 n a is e i e i. Ae s = n a is e i. So, via the natural isomorphism between C n and span{e s e s } n s=1, we see that A and the restriction of à to this subspace are the same operator. In particular, they have the same eigenvalues. This proves that the spectrum of à consists of the spectrum of A together with 0, and hence it is non-negative. Given an undirected graph G we define its adjacency matrix A G to be the Hermitian, strictly off diagonal matrix with an (i, j)-entry of 1 if (i, j) is an edge in G, and 0 otherwise. Define the map δ : M p M p by δ(x) = tr(x) I p, where tr denotes the normalized trace on M p. From here on we study maps γ t,a := t δ + S A where A is an adjacency matrix as just described. It is interesting to ask for what values of t this map will be PPT or entanglement breaking, and to see how the conjecture may be studied within

17 17 this class. An important remark is that for fixed A the sets T P P T and T EB consisting of t such that γ t,a is, respectively, PPT or entanglement breaking are intervals of the form [c, ). First of all we know that γ t,a (I p ) = t I p, so for the map to even be positive we require t to be non-negative. In particular T P P T and T EB have infima, say t P P T and t EB. If we take any t greater than t P P T then the Choi matrix of γ t,a is t t P P T I I +C γtp p, P T,A which is a sum of PPT matrices hence it is PPT. The argument for the EB case is the same, and this lemma follows. Lemma 3.1. For a fixed adjacency matrix A, we have T P P T = [t P P T, ) and T EB = [t EB, ). Remark 3.2. We know that these sets are closed since the sets of PPT and of EB maps are closed, and for any A the function t γ t,a is continuous. A simple computation shows that γt,a 2 = γ t 2,A, so if the PPT-squared conjecture holds then we must always have t 2 P P T t EB. This will indeed be the case, but we must establish sharp enough bounds on these constants to conclude the above inequality. First note that in the case A = 0 we have that t P P T = t EB = 0 so the above inequality trivially holds. Let s assume from now that A 0. We ll first examine the PPT case, and so we must determine restrictions imposed by requiring t p I I +C S A T and t p I I +C S A to be positive, where T is the transpose map M p M p. For the first case, observe that C SA T = (i,j) E E ji E ij, so that (C SA T ) 2 = (i,j) E E ii E jj is a diagonal matrix of only 1 s and 0 s. In particular, the spectrum of (C SA T ) 2 must be a subset of {0, 1}, but then C SA T may only have eigenvalues -1, 0 and 1. So for t = p we will certainly have that γ t,a T is completely positive, and it is minimal exactly when -1 is an eigenvalue of C SA T. In fact, this will always be the case; choose (k, l) such that a k,l = 1 and notice that C SA T (e k e l e l e k ) = e l e k e k e l. As for requiring t p I I +C S A to be positive, we find the minimal eigenvalue of C SA = (i,j) E(G) E ij E ij, where G is the associated graph of A. Notice that C SA is identically zero on the space spanned by e k e l, with k l, and on the span of e k e k it behaves exactly as A acting on C p. Since A is a non-zero Hermitian with zero trace then it must have a negative eigenvalue, and we ll denote its minimal eigenvalue by λ G. We have that λ G I I +C SA T is positive, and it is non-positive for any strictly smaller multiple of I I. The above paragraphs yield the following lemma: Lemma 3.3. If A G is a non-zero adjacency matrix then t P P T = p max{1, λ G }, where λ G is the minimal eigenvalue of A G.

18 18 We would now like to fully understand t EB, although this will pose a greater issue as it is rarely clear when a matrix is algebraically separable in a tensor product. We present here a natural upper bound for t EB via a simple computation. Recall that we view A as being the adjacency matrix of a graph G = (V, E). Lemma 3.4. For an arbitrary adjacency matrix A M p, γ pd,a = pd δ +S A is entanglement breaking, where d denotes the maximum edge degree in G = (V, E). Hence, t eb pd. Proof. We proceed by considering the Choi matrix C Φ of Φ = pd δ + S A and showing it is separable. It is easy to see that C Φ = d I p I p + E ij E ij = D+ E ii E jj +E jj E ii +E ij E ij +E ji E ji, (i,j) E (i,j) E i<j where D is a diagonal matrix consisting of 1 s and 0 s (hence it is separable). So it suffices to show that matrices of the same form as the summand on the right hand side (where i and j may vary from 1 to p) are separable. We use only the four following positive matrices in M p to prove this fact: Q 1,i,j = E i,i + E j,j + E ij + E ji Q 2,i,j = E i,i + E j,j + E ij E ji Q 3,i,j = E i,i + E j,j + ie ij ie ji Q 4,i,j = E i,i + E j,j ie ij + ie ji, where i < j vary from 1 to p. A routine computation shows that 4 ( (E i,i + E j,j ) (E i,i + E j,j ) + E ij E ij + E ji E ji ) = Q 1,i,j Q 1,i,j + Q 2,i,j Q 2,i,j + Q 3,i,j Q 4,i,j + Q 4,i,j Q 3,i,j. Summing over all edges we get that R = 4 (E i,i + E j,j ) (E i,i + E j,j ) + 8 (i,j) E (i,j) E E i,j E j,i, is separable. Now in the sum (i,j) E (E i,i + E j,j ) (E i,i + E j,j ), for k l each term E k,k E l,l appears at most once, while E i,i E i,i occurs exactly 2d i 2d times. Since each term E k,k E l,l is separable, we see that we can add a separable term Q to R so that R + Q = 8d ( ) E i,i E j,j + 8 E i,j E j,i = 8C Φ, i,j (i,j) E and it follows that C Φ is separable so that Φ = γ pd,a is entanglement breaking. Before stating the next result note that the Schur product A B of two adjacency matrices is again an adjacency matrix.

19 Corollary 3.5. If A and B are adjacency matrices and the maps γ t1,a and γ t2,b are PPT, then their composition is entanglement breaking. Proof. The composition evaluates to γ t1 t 2,A B, so if either A or B is zero then the composition is the map X t 1 t 2 tr(x). This map is clearly entanglement breaking. If both are non-zero matrices then t 1 t 2 p 2, and this is necessarily greater than t EB for any adjacency matrix of size p since the degree of any vertex cannot exceed p 1. The corollary follows. 19

20 20 References [1] Christandl, Matthias. Bipartite Entanglement: A Cryptographic point of view. Diss. PhD thesis, University of Cambridge, [2] Gurvits, Leonid. Classical deterministic complexity of Edmonds Problem and quantum entanglement. Proceedings of the thirty-fifth annual ACM symposium on Theory of computing. ACM, [3] Horodecki, Micha, Pawe Horodecki, and Ryszard Horodecki. Separability of n- particle mixed states: necessary and sufficient conditions in terms of linear maps. Physics Letters A (2001): 1-7. [4] Horodecki, Michael, Peter W. Shor, and Mary Beth Ruskai. Entanglement breaking channels. Reviews in Mathematical Physics (2003): [5] Jamiokowski, Andrzej. Linear transformations which preserve trace and positive semidefiniteness of operators. Reports on Mathematical Physics 3.4 (1972): [6] Johnston, Nathaniel, et al. Minimal and maximal operator spaces and operator systems in entanglement theory. Journal of Functional Analysis (2011): [7] Klep, Igor, et al. There are many more positive maps than completely positive maps. arxiv preprint arxiv: (2016). [8] V.I. Paulsen, Completely Bounded Maps and Operator Algebras, Cambridge Studies in Advanced Mathematics 78, Cambridge University Press, [9] Paulsen, Vern. Entanglement and Non-locality. Course notes, University of Waterloo, Winter [10] Watrous, John. Theory of Quantum Information. Chapter 18, [11] Woronowicz, Stanisaw Lech. Positive maps of low dimensional matrix algebras. Reports on Mathematical Physics 10.2 (1976): Department of Pure Mathematics, University of Waterloo, Waterloo, ON, Canada N2L 3G1 address: nmanor@uwaterloo.ca