Supplemental material for : Entangling Capacities and the Geometry of Quantum Operations

We are only concerned with finite-dimensional Hilbert spaces, denoted by H. The symbol L is for the vector space formed by linear mappings, e.g. L(H1,H2) is the space constituted by linear mappings from H1 to H2, L(H) := L(H,H) is a space of operators, and Choi isomorphism T is an isomorphism between L(L(H1),L(H2)) and L(H1)⊗ L(H2) [2]. The term “linear mapping” usually refers to one from an operator algebra to another, i.e. they constitute the space L(L(H1),L(H2)); the context dictates what they actually mean. The symbol L is for a linear mapping, while S and Si are for quantum operations and sub-operations thereof: S and Si are CP. In general we use S without a subscript for a TP (deterministic) quantum operation, but sometimes we’ll be a bit loose on this, as in Sec. 5.1.


Linear mappings and quantum operations
We are only concerned with finite-dimensional Hilbert spaces, denoted by H. The symbol L is for the vector space formed by linear mappings, e.g. L(H 1 , H 2 ) is the space constituted by linear mappings from H 1 to H 2 , L(H) := L(H, H) is a space of operators, and Choi isomorphism T is an isomorphism between L(L(H 1 ), L(H 2 )) and L(H 1 ) ⊗ L(H 2 ) [2].
The term "linear mapping" usually refers to one from an operator algebra to another, i.e. they constitute the space L(L(H 1 ), L(H 2 )); the context dictates what they actually mean. The symbol L is for a linear mapping, while S and S i are for quantum operations and sub-operations thereof: S and S i are CP. In general we use S without a subscript for a TP (deterministic) quantum operation, but sometimes we'll be a bit loose on this, as in Sec. 5.1.

Entangling capacity
Entangling capacity of a deterministic operation or sub-operation S i with respect to an entanglement measure m, when aided by an ancilla, is defined as [3,4,5,6] EC m (S i ) := max ρ {m(I a ⊗ S i (ρ)) − m(ρ)}, where m gives the same result whether the input state is normalized or not. For a probabilistic operation S with sub-operations S i , it's EC m (S) := max ρ i p i m(I a ⊗ S i (ρ)) − m(ρ) .
Here the ancilla space H a can be any finite-dimensional inner-product space, I a is the identity mapping on L(H a ), and the maximizations are over all density operators in L(H a ⊗ H). In Ref. [6], entangling capacity was actually defined to be the supremum, rather than maximum over all states. However, if the measure m is continuous, then since the space is finite-dimensional, physical states in the space of Hermitian operators is a compact set K, and m(K) is also compact [7,8], so there's no difference between the supremum and maximum.

Hilbert-Schmidt inner product
We make use of Hilbert-Schmidt inner product: for O i ∈ L(H i ) linear in the second argument, like bra-ket. It's an inner product, because it satisfies all the properties for an inner product [9].
The trace of an operator after a linear mapping L can be expressed in terms of an inner product between the operator and L † (I).
If L is HP, it always admits an operator-sum representation [11,12,13]: Therefore From this it's apparent L † (I) is Hermitian if L is HP. A simple L † (I) is more elegant and concise than the right side of (7), and that's one of the reasons why we express our results in, for example, S Γ − † (I), rather than the operators constituting the operator-sum representation of S Γ . Note [10] ( We will utilize this relation frequently for proofs.

Orthonormal basis of operators and Choi isomorphism
Consider a Hilbert space H 1 with an orthonormal basis {a i }. With Hilbert-Schmidt inner product, we can choose an orthonormal basis of operators for L(H 1 ) [10,14], by defining Choi isomorphism T then is defined as [12] T (L) := I ⊗ L( It can be found L(O) = tr 1 (O T ⊗ IT (L)) [15]: The subscript beside tr refers to the party to be traced out; in this case, it's H 1 (or L(H)).

Bipartite system and partial transposition
When there are two parties Alice and Bob, let's consider a linear mapping from L(H A1 ⊗H B1 ) to L(H A2 ⊗H B2 ). We don't assume H A1 and H A2 are isomorphic, similarly for H B1 and H B2 , so the dimensions before and after L can be different. Assign a basis for L(H A1 ) by (9), with {|a i } being an orthonormal basis of H A1 ; then define a basis for L(H B1 ): where {|bi } is an orthonormal basis of B 1 . Choi isomorphism for this composite system then is . From now on {E ij } and {F ij } will denote bases of operators. Transposition T is a linear mapping: T : L(H) → L(H). Similarly partial transposition, Γ = T A ⊗ I B , is also a linear mapping Γ : L(H A ⊗ H B ) → L(H A ⊗ H B ). They are both the inverses to themselves, i.e. T • T = I and Γ • Γ = I A ⊗ I B = I AB , so both are isomorphisms.
Partial transposition of a linear mapping L is defined as [16] L Γ := Γ • L • Γ. (13) Note for L : , and the second one maps L(H A2 ⊗ H B2 ) to L(H A2 ⊗ H B2 ). (13) itself is a linear mapping on linear mappings (on operators): If we define f (L) := L Γ , then Furthermore, as (L Γ ) Γ = L, f is its own inverse, so it is also an isomorphism.

Trace, Schatten norm and positive operators
For P 1 , P 2 ≥ 0, by the eigendecomposition P 1 = i p i |ψ i ψ i | we can find [6]: where eigenvalue 0 is included here. The inequality on the right can be regarded as an application of Hölder's inequality [17]. Either side of (14) becomes an equality iff the subspace orthogonal to ker P 2 is in the eigensapce of P 1 with the maximum or minimum eigenvalue, e.g. if P 1 = pI, trP 1 P 2 = p||P 2 || 1 . ker refers to the kernel or null space. More generally, for an operator O its Schatten p-norm [9,18,19] is For p = ∞ it is (defined as) the largest singular value of O, hence the same as the operator norm, || · · · || in our notation; for p = 1 it is exactly the trace norm [18,19]. Matrix Hölder inequality [20] states that for 1 ≤ p, q ≤ ∞ satisfying 1/p + 1/q = 1: As the upper bounds derived in our work are mainly based on (14), we can actually derive more general bounds using (16) and p-and q-norms, with 1/p + 1/q = 1 and 1 ≤ p, q ≤ ∞. Further discussions will be given in Sec. 4.6.
From (14), we can easily see why for a sub-operation S i , otherwise the probability could surpass 1 or be negative, and also why a TP mapping L should have L † (I) = I, which can also be regarded as a special case of the fact that (O 1 |O 2 ) = trO 2 for all O 2 if and only if O 1 = I. Additionally, the composition of TP mappings is still TP, so for a deterministic S S Γ † (I) = I.
For any positive operator P , its trace is equal to its trace norm, trP = ||P || 1 . In this supplemental material (as in the main paper) we will use them interchangeably. As an example, for any CP mapping L, L † (I) ≥ 0, so trL † (I) = ||L † (I)|| 1 . More specifically, S Γ Eigendecomposition H = H + − H − is one of such decompositions. Let ker denote the kernel or null space of an operator and ran the range or image of a mapping. Define similarly for H ± . Hence, and those three subspaces are mutually orthogonal.
Here is a lemma concerning such a decomposition, which is a generalization of Lemma 2 from Ref. [21]: Proof:

Only eigendecomposition minimizes traces
Suppose we want to minimize both tr H ± simultaneously. Because H ± ≥ 0 and because of (22), for both sides of (23) to be equal, i.e. to minimize both tr H ± , H ± must be the same as H ± (as H ± ⊆ H ± ), implying and where |ψ ± and |φ ± are any vectors in H ± ; the same is true for any vectors in ker H. Or alternatively, we can directly apply the spectral theorem [22]: As H ± are orthogonal, (after eigendecomposing H ± ) H + − H − is an orthogonal spectral resolution [22] of H.

How to construct such a decomposition
In this section we will discuss given a Hermitian operator H, how to construct such a decomposition for it. The goal is to find a Hermitian operator H such that From (22) and H ± ⊆ H ± it may seem H ± ≥ H ± , but this is not true in general-what (22) implies is actually H ± ≥ H ± to the restriction of H ± . To be specific, for a vector |ψ + + |ψ − where |ψ ± ∈ H ± , the value of even though ψ ± | H |ψ ± ≥ 0, is not necessarily non-negative, as it depends on ψ ± | H |ψ ∓ as well. In terms of matrices, it just reflects the fact that a matrix that has non-negative diagonal entries are not necessarily positive (semi-definite).
Let's choose a random orthonormal basis {|φ i }, and use it to construct H : The eigenvalues a i are to be determined. Defining matrices M and A as Now we can try different combination of {a i }; if (29) is satisfied, then we find a valid decomposition. Sylvester's theorem may be useful in this process. The benefit of choosing the eigenvectors of H as the standard basis is that we have less parameters to fix. Here are some guidelines on how to select the basis and eigenvalues: 3. As a positive operator is also positive to the restriction of any subspace, it is advisable to determine the negative parts of H first, and check if H ± + H is positive when restricted to the subspace H − ; however it should be emphasized even if H ± + H ≥ 0 to the restriction to this subspace, it does not mean H ± + H ≥ 0 over the entire space. 4. Similarly, for M − A ≥ 0 it is necessary that M − A have non-negative diagonal elements, but it is not a sufficient condition, as we have pointed out. The third and fourth points can be regarded as an application of Sylvester's theorem.

Orthogonality of ensembles
Lemma 2. Suppose P 1 and P 2 are two positive operators, and they have such ensembles: [23,24] where each {|ψ i j } is a set of nonzero vectors that aren't necessarily normalized or mutually orthogonal. Then each eigenvector/eigenspace of P 2 corresponding to a nonzero eigenvalue is orthogonal to each of P 1 if and only if Proof. From Ref. [23], if the eigenensembles of P i are where |e i j aren't normalized, then there exist unitary matrices U i such that and If In other words, because each |ψ i j is a linear combination of {|e i j }, and each |e i j is also linear combinations of {|ψ i j }, orthogonality of two ensembles guarantees that of the other two. Here's a more elegant proof: The orthogonality of the vectors as stated in the lemma is satisfied if and only if their spans are orthogonal. From (32), the spans of {ψ i j } are identical to (ker P i ) ⊥ = ranP i , which of course are also identical to the spans of {e i j }. Therefore the orthogonality of one pair implies that of the other.
3 Adjoint of a linear mapping and complex conjugation of an operator . Hence tr 2 in the following lemma traces out L(H 2 ).
Proof. As in Sec. 1.4, we can find where L(E ij ) * = (L(E ij )) * . In addition, since E * ij = E ij ,

Complex conjugation and basis
Note the importance of the basis for H 1 chosen in the proof above, because the transpose and complex conjugation of an operator depend on the basis. As for H 2 , since tr(A * ) = (trA) * with whichever (orthonormal) basis, it does not matter. Now assume the mapping L is HP, so we can perform a spectral decomposition on T (L). Suppose its eigendecomposition is where {b i } is any orthonormal basis of H 2 , and E ij = |a i a j | is the basis chosen for Choi isomorphism. Then, where We thus obtain and {|v * i } is an orthonormal basis. For a vector |v ∈ H 1 ⊗ H 2 , after fixing the basis for H 1 , |v * depends on the basis for H 2 , but tr 2 |v * v * | would be the same regardless of the basis for H 2 .

For a probabilistic operation composed of sub-operations
The upper bounds of part 2 can be applied to a deterministic operation S = i S i . 3. With an initial negativity E N , the expected negativity, i.e. probability times p i the actual negativity E N i , after a sub-operation S i is bounded by: The entangling capacity of a sub-operation is bounded from below by: All the upper bounds remain the same after the addition of an ancilla:

Some equalities
Here are some relevant equalities. Suppose L : L(H 1 ) → L(H 2 ), and the dimension of H 1 is d.

Upper bounds
We start with Lemma 4. For an HPTP mapping L and Hermitian H,

Proof. For any Hermitian H [21]
With an HP L, since L ± are CP, by Lemma 1 [6] trL Since L is TP as well, (14).
By (45) and L being TP, The proof is completed.

Deterministic operation
An operation S is CP, so it is HP [2,11,12]. Because Γ is HPTP, S Γ is HPTP. Hence we can apply Lemma 4 to S Γ (ρ Γ ) = S(ρ) Γ : and The upper bounds from part 1 of Proposition 1 are proven. For L = i L i , where L i are HP and L is TP, we can choose L ± = i L i± . With by Lemma 4 we obtain by the triangle inequality. Therefore for a deterministic operation and Hence we have proved part 2 of Proposition 1 can be applied to deterministic S = i S i .
Let's emphasize again S Γ ± = i ( S Γ i ± ) are some of the possible decompositions of S Γ , and they may or may not be the ideal choices that yield the smallest bounds.

Probabilistic operation
Let's prove the upper bounds from part 2 of Proposition 1 for probabilistic operations. The average negativity after S is Furthermore, by (46) Essentially, we now have and S is TP as in the deterministic case, c.f. (46). Thus Lemma 4 can be used, and we essentially recover the right sides of (49) for the bounds of i p i E N i − E N , where E N is the initial negativity.
As to the bounds for EC L of a probabilistic operation, with an initial logarithmic negativity E L : by the concavity of logarithm and Jensen's inequality [17,25,26]. By (45), because S Γ = i S Γ i is TP. Now we are back to (52). Following the same steps as before, we can obtain the desired result.

Deterministic operation
Employing the method proposed by Campbell [6], to obtain the lower bounds for entangling capacities, an ancilla is used: We apply the operation I AaBa ⊗ S AsBs on the state |Ψ defined as: where is a maximally entangled state, similarly for |Ψ BaBs . Here A a and B a are the ancilla, and are clones of the original system A s and B s (By clones we mean H Aa and H Ba are isomorphic to H As and H Bs . An ancilla in general does not necessarily have to be isomorphic to the system). (55) is essentially a normalized Choi isomorphism.
We will ignore the subscripts of I AaBa and S AsBs from now on. As A and B are not entangled, alone can give us a lower bound for entangling capacity. With Γ = T A ⊗ I B , we have By Corollary 1, trT (S Γ ± ) = trS Γ ± † (I), so Alternatively, we can directly calculate trT (S Γ ± ) without using Corollary 1: because trL(I) = (L † (I)|I) = tr(L † (I)) † , and L † (I) is Hermitian is L is HP, (7). Hence We can use (45) to acquire the bound for logarithmic negativity as in Proposition 1. Or, because T (S Γ ± ) are orthogonal, by (60) which gives us another expression of the lower bound. We can also use (44) to show this relation.

Sub-operation
To obtain the lower bounds for a sub-operation the procedures are pretty much the same, but now we should take the probability into account: c.f. (60) or (61). Use the same method to obtain trI ⊗ S Γ i − (|Ψ Ψ|). Thus the negativity is Similarly, With (69) we can adjust (66) and (68) to our liking, such as in Proposition 1. Note in general we can't expect S Γ i † (I) ≥ 0, so its trace may not equate to its trace norm.

Probabilistic operation
For the average negativity of a probabilistic operation, by (65) and (66) For the average logarithmic negativity, by (65) and (68) Note for the upper bounds the denominator p i could be removed due to concavity (see (53)), which can't be applied here. Also note by Corollary 1 and (69) there are several expressions for these bounds, just like sub-operations.

Proposition 2
As defined in the main paper, for two operators or linear mappings X and Y where ||L|| p := ||T (L)|| p for any linear mappings L.

Upper bounds with an ancilla
Since L ± are CP, so are I ⊗ L ± , and I ⊗ ( L + − L − ) = I ⊗ L + − I ⊗ L − , which is therefore a decomposition of I ⊗ L as a difference of CP mappings, or in our notation, I ⊗ L ± .
With this in mind, let's show the upper bounds are valid with an ancilla; in particular [6] || where I operates on the ancilla. Namely, the bounds obtained for S i are also applicable to I ⊗ S i and vice versa. Note (I ⊗ S i ) Γ = I ⊗ S Γ i . Unlike in Sec. 4.3, the ancilla can be of any dimension, not merely a clone of the original system.
Here let's prove another corollary of Lemma 3: Proof. Choose {|e i } as the orthonormal basis of the ancilla, and define G ij := |e i e j |. We can find By Lemma 3, (I ⊗ L) † (I) = tr 2 T (I ⊗ L) * . Because G * ij = G ij , The reader can find a discussion on complex conjugation of operators in Sec. 3.2. This proof can be compared with that in Ref. [6].

Upper bounds using Schatten p-norm
All the upper bounds come from where P originates from the adjoint of a CP mapping. By matrix Hölder inequality (16) [20], where 1 ≤ p, q ≤ ∞ and 1/p + 1/q = 1. We do not need to take the absolute value as trP 1 P 2 ≥ 0. For a Hermitian operator H, and the upper bound for logarithmic negativity will be state-dependent when q = 1. Or for (77), we may want to use another norm for states, then the norm for S Γ − (I) can be adjusted accordingly. The rest can be derived easily. [6], it was shown for a deterministic operation S = i S i , where S i (ρ) = V i ρV † i and V i has a Schmidt decomposition [28] V

Equivalence of the bounds
the entangling capacity has an upper bound:

It was shown by Campbell for a unitary operator U
while our lower bound is Here we would like to show , and the lower bound obtained in Ref. [6], (83), is the same as that from Proposition 1, (84).

Upper bound
Utilizing the tricks from Appendix E of Ref. [6], we are able to give the following proof.

Single sub-operation
Let's first consider just one sub-operation S i , and drop the subscript i for this moment. Suppose S(O) = V OV † , with V = i λ i A i ⊗ B i being the Schmidt decomposition. Then Define We can see for i = j As each summation on the right side of (88) is in the Kraus form, each is a CP mapping, so we have Now the problems is, is S Γ ± = S Γ ± ? This is equivalent to T ( S Γ ± ) having orthogonal eigenvectors, by Lemma 1 (and Choi isomorphism, well, being bijective). We can find {V + ij } and {V − kl } are mutually orthogonal: the orthogonality of the rest can be shown by the orthonormality of {A i } and {B i }. Choose an orthonormal basis {|a i } for the composite system A ⊗ B whose dimension is d = d A d B . Like (56), define Then the Choi isomorphism of an operation by a single V ± kl is Therefore T ( S Γ ± ) are ensembles of pure states. It was proved in Appendix F of Ref. [ By Lemma 2, T ( S Γ ± ) have orthogonal eigenvectors, so S Γ ± as defined in (89) are equal to S Γ ± . Therefore,

In the entirety
Let's come back to the operation is the Schmidt decomposition. By (43), for a TP L = i L i with each L i being HP, we have along with (94) we obtain,

PPT and separable unitary operations
Proposition 3. In a finite-dimensional system, the following statements for either a unitary operation S(ρ) = U ρU † or a pure state |ψ are equivalent: 1. The Schmidt rank of U or |ψ is 1.
2. The operation/state is separable.
3. The operation/state is PPT.
That PPT pure states are necessarily separable has already been shown in, e.g. [21,29], where the negativity of a pure state |ψ was explicitly calculated, given its Schmidt decomposition. Below we will provide an alternative proof for pure states.

Proof for pure states
By definition a pure state is separable if and only if its Schmidt rank is 1. Suppose the Schmidt decomposition of a pure state is Since partial transpose with respect to different bases are related unitarily, we can do this with respect to any basis, and let's choose {|a i }. We obtain There's a decomposition of ρ Γ as ρ Γ = ρ Γ As the vectors are orthonormal, (100a) and (100a) together are eigendecomposition of ρ Γ ; in other words ρ Γ ± = ρ Γ ± . Therefore, ρ Γ − = 0, i.e. ρ is PPT if and only if the Schmidt rank of |ψ is 1.

Proof for unitary operations
If U is separable, then the state above is separable with respect to A and B. Because of (93), for ( i λ i ) 2 to become d A d B the Schmidt rank of U must be 1, i.e. separable.

Schmidt decomposition and eigendecomposition
The method from Ref. [6] used to show Lemma 5 and Proposition 3 can be further generalized as follows.
As was shown in Sec. 5.1, for a (sub-)operation S(A) = V AV † where the Schmidt decomposition of V is Because both {A i } and {B i } are composed of orthonormal operators, (V ± ij |V ± kl ) = δ ik δ jl , and thus {V + ij : i ≤ j} and {V − ij : i < j} are sets of orthonormal operators, and V + ij ⊥ V − kl . Therefore after Choi isomorphism, T (S Γ ) becomes an ensemble of orthogonal pure states, c.f. (92) and (93), giving us an eigendecomposition of it.

2 ⊗ 3 unitary operation
For the upper and lower bound to be the same with S Γ − = S Γ − , we need S Γ † − (I) ∝ I, i.e. the maximum and minimum eigenvalues of S Γ † − (I) ∝ I should be identical. From Fig. 1 they seem to be the same at 1/2 when α = 2π/3 and β = 0, which can indeed be verified analytically, so the exact entangling capacity at α = 2π/3 and β = 0 can be acquired.

Degeneracy
When α = 2π/3 and β = 0, the spectrum of T (S Γ ) is {−3, 3, 3, 3}, which has degeneracy. When trying to find the optimal state for it i.e. to solve (120), the eigenvectors with eigenvalue 3 that came out of the program were not orthogonal but were linearly independent, and we did not try to orthogonalize them for the following reason: The dimension of an eigenspace with a degenerate eigenvalue is higher than one. Suppose the said eigenspace can be spanned by {|v i } and {| v i }, both with the same cardinality but the former being an orthonormal set and the latter being normalized but not orthogonal. Hence one is the linear combination of the other: |v i = j c ij | v j , c ij is non-singular. The Kraus operator corresponding to an eigenvector |v i = jk d i jk |a j |b k of a linear mapping is which is a linear process, that is, V i = j c ij V j , where V j := V (| v j ).
Therefore, V i |ψ = j c ij V j |ψ , and {V i |ψ } has the same span as { V i |ψ } does. For two (sub)spaces to be orthogonal, it does not matter which vectors we choose to span them, just as discussed in the proof for Lemma 2. This explains why there is no need to orthogonalize eigenvectors with the same eigenvalue if we just want to find the optimal state.