A Triality Pattern in Entanglement Theory

Cariello, Daniel

doi:10.2478/qic-2024-0003

Full Article

1.

Introduction

The separability problem in quantum theory [1] asks for a criterion to distinguish the separable states from the entangled states. It is known that separable states in ℳ_k ⊗ ℳ_m form a subset of the positive under partial transpose states (PPT states) and in low dimensions, km ≤ 6, these two sets coincide [2,3] solving the problem. However, in larger dimensions, km > 6, there are entangled PPT states. In addition, for k, m arbitrary, this problem is known to be a hard problem [4], thus any reduction of this problem to a subset of PPT states of ℳ_k ⊗ ℳ_m is certainly important.

In [5], a procedure to reduce the separability problem to a proper subset of PPT states was presented. The idea behind this reduction can be summarized as follows. Let $A = \sum_{i = 1}^{n} A_{i} \otimes B_{i} \in M_{k} \otimes M_{m}$ A = \sum\nolimits_{i = 1}^n {A_i} \otimes {B_i} \in {{\cal M}_k} \otimes {{\cal M}_m} be a quantum state and consider G_A : ℳ_k → ℳ_m and F_A : ℳ_m → ℳ_k as $G_{A} (X) = \sum_{i = 1}^{n} tr (A_{i} X) B_{i}$ {G_A}(X) = \sum\nolimits_{i = 1}^n tr({A_i}X){B_i} and $F_{A} (X) = \sum_{i = 1}^{n} tr (B_{i} X) A_{i}$ {F_A}(X) = \sum\nolimits_{i = 1}^n tr({B_i}X){A_i} , where tr(X) stands for the trace of X.

Now, if A is PPT and X is a positive semidefinite Hermitian eigenvector of F_A ∘ G_A : ℳ_k → ℳ_k, then A decomposes as a sum of states with orthogonal local supports [5, Lemma 8], i.e., $A = (V \otimes W) A (V \otimes W) + (V^{⊥} \otimes W^{⊥}) A (V^{⊥} \otimes W^{⊥}),$ A = (V \otimes W)A(V \otimes W) + ({V^ \bot } \otimes {W^ \bot })A({V^ \bot } \otimes {W^ \bot }), where V, W, V^⊥, W^⊥ are orthogonal projections onto Im(X), Im(G_A(X)), ker(X) and ker(G_A(X)), respectively.

Then the algorithm proceeds to decompose (V ⊗ W)A(V ⊗ W ) and (V^⊥ ⊗ W^⊥)A(V^⊥ ⊗ W^⊥), since they are also PPT, whenever such X is found. Eventually this process stops with $A = \sum_{i = 1}^{n} (V_{i} \otimes W_{i}) A (V_{i} \otimes W_{i}),$ A = \sum\nolimits_{i = 1}^n ({V_i} \otimes {W_i})A({V_i} \otimes {W_i}), where (V_i⊗W_i)A(V_i⊗W_i) cannot be further decomposed for 1 ≤ i ≤ n. These states are named weakly irreducible. Finally, A is separable if and only if each (V_i ⊗ W_i)A(V_i ⊗ W_i) is separable, therefore this algorithm reduces the separability problem to the weakly irreducible PPT case.

Positive under partial transpose states are not the only type of states on which this procedure works because the key feature of this procedure, which is A ∈ ℳ_k ⊗ ℳ_m breaks whenever a certain positive semidefinite Hermitian eigenvector is found, is also true for two other types of quantum states. From now on we say that A ∈ ℳ_k ⊗ ℳ_m has the complete reducibility property, if for every positive semidefinite Hermitian eigenvector X of F_A ∘ G_A : ℳ_k → ℳ_k, we have $A = (V \otimes W) A (V \otimes W) + (V^{⊥} \otimes W^{⊥}) A (V^{⊥} \otimes W^{⊥}),$ A = (V \otimes W)A(V \otimes W) + ({V^ \bot } \otimes {W^ \bot })A({V^ \bot } \otimes {W^ \bot }), where V, W, V^⊥, W^⊥ are orthogonal projections onto Im(X), Im(G_A(X)), ker(X) and ker(G_A(X)), respectively. In [5], a search for other types of quantum states satisfying this property was conducted finding only three types of quantum states: positive under partial transpose states (PPT states), symmetric with positive coefficients states (SPC states) and invariant under realignment states. So far this property has been only verified for this triad of quantum states.

It is not hard to compute a single positive semidefinite Hermitian eigenvector for F_A ∘ G_A : ℳ_k → ℳ_k, when A is positive semidefinite. In the next section, we shall see that that under this hypothesis G_A : ℳ_k → ℳ_m and F_A : ℳ_m → ℳ_k are adjoint positive maps. So if λ is the largest positive eigenvalue of F_A ∘ G_A, then $P = {lim}_{n \to \infty} {(\frac{F_{A} \circ G_{A}}{λ})}^{n}$ P = \mathop {\lim }\nolimits_{n \to \infty } {\left( {{{{F_A} \circ {G_A}} \over \lambda }} \right)^n} is the orthogonal projection onto the eigenspace of F_A ∘ G_A associated to λ. As a limit of positive maps, P is also a positive map. Therefore P (Id) is a positive semidefinite Hermitian matrix belonging to the eigenspace of F_A ∘ G_A associated to λ.

Now, the number of times that A breaks into weakly irreducible pieces is maximized when the non-null eigenvalues of F_A ∘ G_A : ℳ_k → ℳ_k are equal, because in this case we are able to produce many positive semidefinite Hermitian eigenvectors. In this situation, we have the following theorem:

If the non-null eigenvalues of F_A ∘ G_A are equal, then A is separable, when A is PPT or SPC or invariant under realignment [5, Proposition 15]. Notice that the eigenvalues of F_A ∘ G_A are the square of the Schmidt coefficients of A.

In [5, Lemma 42], it was also noticed that every SPC state and every invariant under realignment state is PPT in ℳ₂ ⊗ ℳ₂. Before that in [6], it was proved that a state supported on the symmetric subspace of ℂ^k ⊗ ℂ^k is PPT if and only if it is SPC. This is plenty of evidence of how linked these three types of quantum states are.

In this work we prove new results for this triad of quantum states and a new consequence for their complete reducibility property. One of these results concerns the separability of these states. The aforementioned connections along with our new results lead us to notice a certain triality pattern:

For each proven theorem for one of these three types of states, there are corresponding counterparts for the other two.

We believe that a solution to the separability problem for SPC states or invariant under realignment states would shed light on bound entanglement. This is our reason for studying these types of states.

Next, we would like to point out the origin of these connections which is also the source of the main tools used in this work. For this, we must consider the group of linear contractions [7] and some of its properties. For each permutation σ ∈ S₄, the linear transformation L_σ : ℳ_k ⊗ ℳ_k → ℳ_k ⊗ ℳ_k satisfying $L_{σ} (v_{1} v_{2}^{t} \otimes v_{3} v_{4}^{t}) = v_{σ (1)} v_{σ (2)}^{t} \otimes v_{σ (3)} v_{σ (4)}^{t},$ {L_\sigma }({v_1}v_2^t \otimes {v_3}v_4^t) = {v_{\sigma (1)}}v_{\sigma (2)}^t \otimes {v_{\sigma (3)}}v_{\sigma (4)}^t, where v_i ∈ ℂ^k is called a linear contraction.

The term contraction comes from the fact that ‖L_σ(γ)‖₁ ≤ ‖γ‖₁, for every σ ∈ S₄, whenever γ ∈ ℳ_k ⊗ ℳ_k is a separable state and ‖ ⋅ ‖₁ is the trace norm of a matrix (i.e., the sum of its singular values). Hence, if γ ∈ ℳ_k ⊗ ℳ_kis a state and ‖L_σ(γ)‖₁ > ‖γ‖₁ for some σ ∈ S₄, then γ ∈ ℳ_k ⊗ ℳ_k is entangled. This observation provides two useful criteria for entanglement detection. In cycle notation, they are:

the PPT criterion [2,3], when σ = (34), and
the CCNR criterion [8,9], when σ = (23).

Despite the name – contraction – these linear maps are in fact isometries, as they map an orthonormal basis { ${e_{i} e_{j}^{t} \otimes e_{k} e_{l}^{t}, 1 \leq i, j, k, l \leq k}$ \{ {e_i}e_j^t \otimes {e_k}e_l^t,\;1 \le i,j,k,l \le k\} of ℳ_k ⊗ℳ_k onto itself. Hence, they preserve the Frobenius norm of γ ∈ ℳ_k ⊗ ℳ_k, denoted here by ‖γ‖₂. This norm is an invariant exploited several times in this work.

In addition, the set of linear contractions is a group under composition generated by the partial transposes (L₍₃₄₎(γ) = γ^Γ and L₍₁₂₎) and the realignment map (L₍₂₃₎(γ) = ℛ(γ)). The relations among the elements of this group are extremely useful as we shall see in the proofs of our novel results.

Another very useful fact about the realignment map is that this map is a homomorphism with respect to the usual matrix product and a new product which generalizes the Hadamard product (see item 8) of Lemma 1).

Finally, these maps allow connecting this triad of quantum states from their origins, that is, their definitions:

PPT states are the states that remain positive under partial transpose (γ ≥ 0, γ^Γ ≥ 0) [2,3],
SPC states are the states that remain positive under partial transpose composed with realignment (γ ≥ 0, ℛ(γ^Γ) ≥ 0) ([5, corollary 25] and [5, definition 17]),
invariant under realignment states are the states that remain the same under realignment (γ ≥ 0, ℛ(γ) = γ).

These observations on the group of linear contractions are used throughout this paper; they are the keys to obtain our novel results. Now, let us describe these results.

Our first result is an upper bound on the spectral radius for this triad of states. We show that if $γ = \sum_{i = 1}^{m} C_{i} \otimes D_{i} \in M_{k} \otimes M_{k}$ \gamma = \sum\nolimits_{i = 1}^m {C_i} \otimes {D_i} \in {{\cal M}_k} \otimes {{\cal M}_k} is PPT or SPC or invariant under realignment, then ${∥ γ ∥}_{\infty} \leq min {{∥ γ_{A} ∥}_{\infty}, {∥ γ_{B} ∥}_{\infty}, {∥ R (γ) ∥}_{\infty}},$ {\left\| \gamma \right\|_\infty } \le \min \{ {\left\| {\gamma _A}\right\|_\infty },{\left\| {\gamma _B}\right\|_\infty },{\left\| {\cal R}(\gamma )\right\|_\infty }\} , where ‖.‖_∞ is the operator norm, and $γ_{A} = \sum_{i = 1}^{m} C_{i} tr (D_{i})$ {\gamma _A} = \sum\nolimits_{i = 1}^m {C_i}tr({D_i}) and $γ_{B} = \sum_{i = 1}^{m} D_{i} tr (C_{i})$ {\gamma _B} = \sum\nolimits_{i = 1}^m {D_i}tr({C_i}) are the reduced states. Let us say that the ranks of γ_A and γ_B are the reduced ranks of γ.

Our second result regards the filter normal form of SPC states, invariant under realignment states and certain types of PPT states. This normal form has been used in entanglement theory, for example, to provide a different solution for the separability problem in ℳ₂ ⊗ ℳ₂ [10,11,12] or to prove the equivalence of some criteria for entanglement detection [13]. Here we show that states which are SPC or invariant under realignment can be put in the filter normal form and their normal forms can still be chosen to be SPC and invariant under realignment, respectively. In other words, if A, B ∈ ℳ_k ⊗ ℳ_k are states such that

(1)
A is SPC then there is an invertible matrix R ∈ ℳ_k such that $(R \otimes R) A {(R \otimes R)}^{*} = \sum_{i = 1}^{n} λ_{i} δ_{i} \otimes δ_{i}$ (R \otimes R)A{(R \otimes R)^*} = \sum\nolimits_{i = 1}^n {\lambda _i}{\delta _i} \otimes {\delta _i} , where $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} , $δ_{1} = \frac{Id}{\sqrt{k}}$ {\delta _1} = {{Id} \over {\sqrt k }} , λ_i > 0 and δ_i is Hermitian for every i, and tr(δ_iδ_j) = 0 for i ≠ j.
(2)
B is invariant under realignment then there is an invertible matrix R ∈ ℳ_k such that $(R \otimes \bar{R}) B {(R \otimes \bar{R})}^{*} = \sum_{i = 1}^{n} λ_{i} δ_{i} \otimes \bar{δ_{i}}$ (R \otimes \overline R )B{(R \otimes \overline R )^*} = \sum\nolimits_{i = 1}^n {\lambda _i}{\delta _i} \otimes \overline {{\delta _i}} , where $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} , $δ_{1} = \frac{Id}{\sqrt{k}}$ {\delta _1} = {{Id} \over {\sqrt k }} , λ_i > 0 and δ_i is Hermitian for every i, and tr(δ_iδ_j) = 0 for i ≠ j.

Our PPT counterpart to these two results, Theorem 3, says that every PPT state whose rank is equal to its two reduced ranks can be put in the filter normal form, which we believe is a novel contribution to the filter normal form literature.

We are not aware of any general algorithm capable of determining whether a given state can be put in the filter normal form or not. From this point of view, these results are not only novel but are highly relevant to the literature on the filter normal form.

Our final result is a lower bound for the ranks of these three types of states. We show that the rank of PPT states, SPC states and invariant under realignment states cannot be inferior to their reduced ranks (when they are equal) and whenever this minimum is attained the states are separable.

In [14], it was proved that a state γ such that rank(γ) ≤ max{rank(γ_A), rank(γ_B)} is separable by just being positive under partial transpose. In [15], it was shown that the rank of a separable state is greater or equal to its reduced ranks. So if γ is PPT and rank(γ) ≤ max{rank(γ_A), rank(γ_B)}, then it is separable and rank(γ) = max{rank(γ_A), rank(γ_B)}.

Thus, our last result is known for PPT states, but it is original for SPC states and invariant under realignment states. The proofs presented here for these results are completely original. We use the facts that every PPT state, SPC state and invariant under realignment state with minimal rank can be put in the filter normal form together with the complete reducibility property to finish the proofs. These results show a nice interplay between the filter normal form and the complete reducibility property.

It is quite surprising that all three of these types of states guarantee separability of their states under such condition. In the entanglement literature, the opposite is usually the case.

For example, it is known that any given bipartite state γ satisfies $rank (γ) \geq \frac{max {rank (γ_{A}), rank (γ_{B})}}{min {rank (γ_{A}), rank (γ_{B})}}$ {\rm{rank}}(\gamma ) \ge {{\max \{ {\rm{rank}}({\gamma _A}),{\rm{rank}}({\gamma _B})\} } \over {\min \{ {\rm{rank}}({\gamma _A}),{\rm{rank}}({\gamma _B})\} }} (see [16, Theorem 3.3]) and, whenever the equality holds, this γ becomes entangled (see [17, Theorem 1]). From this point of view, states that reach the minimum possible rank value are usually entangled.

The results described above show how fundamental the complete reducibility property is to entanglement theory, it acts as a unifying approach for many results. Another recent one on entanglement breaking Markovian dynamics was described in [18]. In fact, even outside entanglement theory, we can find consequences of that property, for example, a new proof of Weiner’s theorem [19] on mutually unbiased bases found in [5].

The triality pattern described above together with the complete reducibility property form a potential source of information for entanglement theory.

This paper is organized as follows. In Section 2, we present some preliminary results, which are mainly facts about the group of linear contractions. In Section 3, we obtain an upper bound for the spectral radius of our special triad of quantum states. In Section 4, we show that SPC states and invariant under realignment states can be put in the filter normal form and their normal forms retain their shape. In addition, we show that PPT states with minimal rank can also be put in the filter normal form. In Section 5, we prove that the ranks of our triad of quantum states cannot be inferior to their reduced ranks and whenever this minimum is attained the states are separable.

2.

Preliminary Results

In this section we present some preliminary results. We begin by noticing that G_A : ℳ_k → ℳ_m and F_A : ℳ_m → ℳ_k defined in the introduction are adjoint maps with respect to the trace inner product, when A is Hermitian. The reason behind this is quite simple: If A is Hermitian, then F_A(Y )^* = F_A^*(Y^*) = F_A(Y^*), for every Y ∈ ℳ_m, hence $tr (G_{A} (X) Y^{*}) = tr (A (X \otimes Y^{*})) = tr ({XF}_{A} (Y^{*})) = tr ({XF}_{A} {(Y)}^{*}) .$ tr({G_A}(X){Y^*}) = tr(A(X \otimes {Y^*})) = tr(X{F_A}({Y^*})) = tr(X{F_A}{(Y)^*}).

Notice that for positive semidefinite Hermitian matrices X ∈ ℳ_k, Y ∈ ℳ_m and A ∈ ℳ_k ⊗ ℳ_m, we have tr(A(X ⊗ Y^*)) ≥ 0. Thus, the equality above also shows that G_A and F_A are positive maps (Definition 2) when A is positive semidefinite. So in this case F_A ∘ G_A : ℳ_k → ℳ_k is a self-adjoint positive map.

These maps G_A : ℳ_k → ℳ_m and F_A : ℳ_m → ℳ_k are connected to the following generalization of the Hadamard product extensively used in [20] and required here a few times.

Definition 1

(Generalization of the Hadamard product). Let $γ = \sum_{i = 1}^{n} A_{i} \otimes B_{i} \in M_{m} \otimes M_{k}$ \gamma = \sum\nolimits_{i = 1}^n {A_i} \otimes {B_i} \in {{\cal M}_m} \otimes {M_k} , $δ = \sum_{j = 1}^{l} C_{j} \otimes D_{j} \in M_{k} \otimes M_{s}$ \delta = \sum\nolimits_{j = 1}^l {C_j} \otimes {D_j} \in {{\cal M}_k} \otimes {M_s} . Define the product γ * δ ∈ ℳ_m ⊗ ℳ_s as $γ * δ = \sum_{i = 1}^{n} \sum_{j = 1}^{l} A_{i} \otimes D_{j} tr (B_{i} C_{j}^{t}) .$ \gamma *\delta = \sum\limits_{i = 1}^n \sum\limits_{j = 1}^l {A_i} \otimes {D_j}tr({B_i}C_j^t).

Let us recall some facts regarding this product.

Remark 1

Let $u = \sum_{i = 1}^{k} e_{i} \otimes e_{i} \in ℂ^{k} \otimes ℂ^{k}$ u=\sum\nolimits_{i=1}^{k}{}{{e}_{i}}\otimes {{e}_{i}}\in {{\mathbb{C}}^{k}}\otimes {{\mathbb{C}}^{k}} , where e₁, …, e_k is the canonical basis of ℂ^k.

a)
Notice that $u^{t} (B_{i} \otimes C_{j}) u = tr ((B_{i} \otimes C_{j}) {uu}^{t}) = tr (B_{i} C_{j}^{t})$ {u^t}({B_i} \otimes {C_j})u = tr(({B_i} \otimes {C_j})u{u^t}) = tr({B_i}C_j^t) , where B_i, C_j ∈ ℳ_k. Therefore, $γ * δ = ({Id}_{m \times m} \otimes u^{t} \otimes {Id}_{s \times s}) (γ \otimes δ) ({Id}_{m \times m} \otimes u \otimes {Id}_{s \times s}),$ \gamma *\delta = (I{d_{m \times m}} \otimes {u^t} \otimes I{d_{s \times s}})(\gamma \otimes \delta )(I{d_{m \times m}} \otimes u \otimes I{d_{s \times s}}), which implies that γ * δ is positive semidefinite, whenever γ, δ are positive semidefinite. In addition, $tr (γ * δ) = tr (γ \otimes δ (Id \otimes {uu}^{t} \otimes Id)) = tr (γ_{B} \otimes δ_{A} {uu}^{t}) = tr (γ_{B} δ_{A}^{t})$ tr(\gamma *\delta ) = tr(\gamma \otimes \delta \;(Id \otimes u{u^t} \otimes Id)) = tr({\gamma _B} \otimes {\delta _A}\;u{u^t}) = tr({\gamma _B}\delta _A^t) .
b)
By [20, Proposition 8], γ * δ = (F_γ ((⋅)^t) ⊗ Id)(δ) = (Id ⊗ G_δ((⋅)^t))(γ).
c)
Let F = (uu^t)^Γ ∈ ℳ_k ⊗ ℳ_k and notice that if $γ = \sum_{i = 1}^{n} A_{i} \otimes B_{i} \in M_{k} \otimes M_{k}$ \gamma = \sum\nolimits_{i = 1}^n {A_i} \otimes {B_i} \in {{\cal M}_k} \otimes {{\cal M}_k} , then $F γ F = \sum_{i = 1}^{n} B_{i} \otimes A_{i}$ F\gamma F = \sum\nolimits_{i = 1}^n {B_i} \otimes {A_i} . This real symmetric matrix F is a unitary matrix and it is usually called the flip operator.

Now, notice that $γ * (F γ^{t} F) = \sum_{j} \sum_{i} A_{i} \otimes A_{j}^{t} tr (B_{i} B_{j})$ \gamma *(F{\gamma ^t}F) = \sum\nolimits_j \sum\nolimits_i {A_i} \otimes A_j^ttr({B_i}{B_j}) . Hence $\begin{matrix} tr (γ * (F γ^{t} F)) = tr ((\sum_{i} tr (A_{i}) B_{i}) (\sum_{j} tr (A_{j}) B_{j})) = tr (γ_{B}^{2}) and \\ G_{γ * (F γ^{t} F)} (X) = \sum_{j} \sum_{i} tr (A_{i} X) tr (B_{i} B_{j}) A_{j}^{t} = \sum_{j} tr ((\sum_{i} tr (A_{i} X) B_{i}) B_{j}) A_{j}^{t} = F_{γ} {(G_{γ} (X))}^{t} . \end{matrix}$ \matrix{ {tr(\gamma *(F{\gamma ^t}F)) = tr\left( {\left( {\sum\nolimits_i tr({A_i}){B_i}} \right)\left( {\sum\nolimits_j tr({A_j}){B_j}} \right)} \right) = tr(\gamma _B^2)\;and} \cr {{G_{\gamma *(F{\gamma ^t}F)}}(X) = \sum\nolimits_j \sum\nolimits_i tr({A_i}X)tr({B_i}{B_j})A_j^t = \sum\nolimits_j tr\left( {\left( {\sum\nolimits_i tr({A_i}X){B_i}} \right){B_j}} \right)A_j^t = {F_\gamma }{{({G_\gamma }(X))}^t}.} \cr }

Next, we discuss some facts about the group of linear contractions. Actually, we need to focus only on three of these maps. The linear contractions important to us are $L_{(34)} (γ) = γ^{Γ}, L_{(24)} (γ) = γ F and L_{(23)} (γ) = R (γ) .$ {L_{(34)}}(\gamma ) = {\gamma ^\Gamma },{L_{(24)}}(\gamma ) = \gamma F\;{\rm{and}}\;{L_{(23)}}(\gamma ) = {\cal R}(\gamma ).

Below we discuss several properties of these linear contractions such as relations among these elements and how they behave with respect to the product defined in Definition 1.

Lemma 1

Let γ, δ ∈ ℳ_k ⊗ ℳ_k and v = Σ_i a_i ⊗ b_i, w = Σ_j c_j ⊗ d_j ∈ ℂ^k ⊗ ℂ^k.

(1)
ℛ(vw^t) = V ⊗ W, where $V = \sum_{i} a_{i} b_{i}^{t}$ V = \sum\nolimits_i {a_i}b_i^t , $W = \sum_{j} c_{j} d_{j}^{t}$ W = \sum\nolimits_j {c_j}d_j^t .
(2)
ℛ(ℛ(γ)) = γ
(3)
ℛ((V ⊗ W )γ(M ⊗ N )) = (V ⊗ M^t)ℛ(γ)(W^t ⊗ N)
(4)
ℛ(γF )F = γ^Γ
(5)
ℛ(γ^Γ) = ℛ(γ)F
(6)
ℛ(γF ) = ℛ(γ)^Γ
(7)
ℛ(γ^Γ)^Γ = γF
(8)
ℛ(γ * δ) = ℛ(γ)ℛ(δ) (i.e., ℛ is a homomorphism).
(9)
$R (F \bar{γ} F) = R {(γ)}^{*}$ {\cal R}(F\overline \gamma F) = {\cal R}{(\gamma )^*}

Proof

Items (1–6) were proved in items (2–7) of [5, Lemma 23]. For the other three items, we just need to prove them when γ = ab^t ⊗ cd^t and δ = ef^t ⊗ gh^t, where a, b, c, d, e, f, g, h ∈ ℂ^k.

Item (7): ℛ(ab^t ⊗ cd^t)^Γ)^Γ = ℛ(ab^t ⊗ dc^t)^Γ = (ad^t ⊗ bc^t)^Γ = ad^t ⊗ cb^t.

Now, (ab^t ⊗ cd^t)F = ad^t ⊗ cb^t. So ℛ(γ^Γ)^Γ = γF.

Item (8): ℛ(ab^t ⊗ cd^t * ef^t ⊗ gh^t) = ℛ(ab^t ⊗ gh^t)(d^tf)(c^te) = (ag^t ⊗ bh^t)(d^tf)(c^te).

Now, ℛ(ab^t ⊗ cd^t)ℛ(ef^t ⊗ gh^t) = (ac^t ⊗ bd^t)(eg^t ⊗ fh^t) = (ag^t ⊗ bh^t)(c^te)(d^tf).

So ℛ(γ * δ) = ℛ(γ)ℛ(δ)

Item (9): $R (F \bar{a} {\bar{b}}^{t} \otimes \bar{c} {\bar{d}}^{t} F) = R (\bar{c} {\bar{d}}^{t} \otimes \bar{a} {\bar{b}}^{t}) = \bar{c} {\bar{a}}^{t} \otimes \bar{d} {\bar{b}}^{t}$ {\cal R}(F\overline a {\overline b ^t} \otimes \overline c {\overline d ^t}F) = {\cal R}(\overline c {\overline d ^t} \otimes \overline a {\overline b ^t}) = \overline c {\overline a ^t} \otimes \overline d {\overline b ^t}

Now, $R {({ab}^{t} \otimes {cd}^{t})}^{*} = {({ac}^{t} \otimes {bd}^{t})}^{*} = \bar{c} {\bar{a}}^{t} \otimes \bar{d} {\bar{b}}^{t}$ {\cal R}{(a{b^t} \otimes c{d^t})^*} = {(a{c^t} \otimes b{d^t})^*} = \overline c {\overline a ^t} \otimes \overline d {\overline b ^t} .

So $R (F \bar{γ} F) = R {(γ)}^{*}$ {\cal R}(F\overline \gamma F) = {\cal R}{(\gamma )^*} .

The next lemma is important for Theorem 3, it says something very interesting about PPT states which remain PPT, under realignment: They must be invariant under realignment.

Lemma 2

Let γ ∈ ℳ_k ⊗ ℳ_k be a positive semidefinite Hermitian matrix. If γ and ℛ(γ) are PPT, then γ = ℛ(γ).

Proof

By item (4) of Lemma 1, γ^Γ = ℛ(γF )F.

Now, since F² = Id, γ^ΓF = ℛ(γF ).

Next, by item (6) of Lemma 1, ℛ(γF ) = ℛ(γ)^Γ. So γ^ΓF = ℛ(γ)^Γ is a positive semidefinite Hermitian matrix by hypothesis.

Since F, γ^Γ and γ^ΓF are Hermitian matrices, γ^ΓF = Fγ^Γ. So there is an orthonormal basis of ℂ^k ⊗ ℂ^k formed by symmetric and anti-symmetric eigenvectors of γ^Γ.

Remind that γ^Γ and γ^ΓF are positive semidefinite, hence γ^Γ = γ^ΓF.

Finally, we have noticed that γ^ΓF = ℛ(γ)^Γ. So γ^Γ = ℛ(γ)^Γ, which implies γ = ℛ(γ).

The next lemma is used in this work a few times. Although simple, we state it here in order to better organize our arguments.

Lemma 3

Let γ ∈ ℳ_k ⊗ ℳ_k. The singular values of the linear map G_γ : ℳ_k → ℳ_k and the matrix ℛ(γ) ∈ ℳ_k ⊗ ℳ_k are equal and their largest singular values can be computed as ${∥ G_{γ} ∥}_{\infty} = {∥ R (γ) ∥}_{\infty} = max {| tr (γ (X \otimes Y)) |, {∥ X ∥}_{2} = {∥ Y ∥}_{2} = 1} .$ {\left\| {G_\gamma }\right\|_\infty } = {\left\| {\cal R}(\gamma )\right\|_\infty } = \max \{ |tr(\gamma (X \otimes Y))|,\;{\left\| X\right\|_2} = {\left\| Y\right\|_2} = 1\} .

In addition, if γ is Hermitian, then there are Hermitian matrices γ₁, δ₁ ∈ ℳ_k such that ‖ℛ(γ)‖_∞ = tr(γ(γ₁ ⊗ δ₁)), where ‖γ₁‖₂ = ‖δ₁‖₂ = 1.

Proof

It was proved in item 1) of [5, Lemma 23] that $V \circ F_{γ^{*}} \circ V^{*} = R ({(γ^{*})}^{Γ}),$ V \circ {F_{{\gamma ^*}}} \circ {V^*} = {\cal R}({({\gamma ^*})^\Gamma }), where V : ℳ_k → ℂ^k ⊗ ℂ^k is the vectorization map $V (\sum_{i} a_{i} b_{i}^{t}) = \sum_{i} a_{i} \otimes b_{i}$ V\left( {\sum\nolimits_i {a_i}b_i^t} \right) = \sum\nolimits_i {a_i} \otimes {b_i} . Since the vectorization map is an isometry, the singular values of F_γ^*: ℳ_k → ℳ_k and ℛ((γ^*)^Γ) ∈ ℳ_k ⊗ ℳ_k are equal.

Now, by item 5) of Lemma 1, ℛ((γ^*)^Γ) = ℛ(γ^*)F, where F is the flip operator, which is unitary. Hence the singular values of ℛ((γ^*)^Γ) and ℛ(γ^*) are the same.

Next, items 2) and 9) of Lemma 1 imply that $R (γ^{*}) = F \bar{R (γ)} F$ {\cal R}({\gamma ^*}) = F\overline {{\cal R}(\gamma )} F . Again, since F is unitary ℛ(γ^*) and ℛ(γ) share their singular values. Hence the singular values of F_γ^*: ℳ_k → ℳ_k and ℛ(γ) ∈ ℳ_k ⊗ ℳ_k are equal.

In addition, G_γ : ℳ_k → ℳ_k is the adjoint of F_γ^*: ℳ_k → ℳ_k, thus G_γ and F_γ^* have the same singular values. Therefore, the singular values of the linear map G_γ : ℳ_k → ℳ_k and the matrix ℛ(γ) ∈ ℳ_k ⊗ ℳ_k are equal. Thus, ‖G_γ ‖_∞ = ‖ℛ(γ)‖_∞.

In order to show that their largest singular values can be computed as stated above, recall that the largest singular value of G_γ : ℳ_k → ℳ_k is equal to $\begin{array}{l} max {{∥ (G_{γ} (W) ∥}_{2}, {∥ W ∥}_{2} = 1} = max {| tr (G_{γ} (W) V) |, W, V \in M_{k} and {∥ W ∥}_{2} = {∥ V ∥}_{2} = 1}, \\ = max {| tr (γ (W \otimes V)) |, W, V \in M_{k} and {∥ W ∥}_{2} = {∥ V ∥}_{2} = 1} . \end{array}$ \matrix{ {\max \{ {\left\| ({G_\gamma }(W)\right\|_2},\;{\left\| W\right\|_2} = 1\} = \max \{ |tr({G_\gamma }(W)V)|,\;W,V \in {{\cal M}_k}\;{\rm and}\;{\left\| W\right\|_2} = {\left\| V\right\|_2} = 1\} ,} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \max \{ |tr(\gamma (W \otimes V))|,\;W,V \in {{\cal M}_k}\;{\rm and}\;{\left\| W\right\|_2} = {\left\| V\right\|_2} = 1\} .} \hfill \cr }

This proves the first part of the lemma. Now for the second part, if γ is Hermitian, then the set of Hermitian matrices is left invariant by G_γ : ℳ_k → ℳ_k. Therefore, there is an Hermitian matrix γ₁ ∈ ℳ_k such that ‖γ₁‖₂ = 1 and ‖G_γ(γ₁)‖₂ = the largest singular value of G_γ.

Notice that ‖G_γ (γ₁)‖₂ = tr(G_γ (γ₁)δ₁) = tr(γ(γ₁ ⊗ δ₁)), where δ₁ = G_γ (γ₁)/‖G_γ (γ₁)‖₂.

So ‖ℛ(γ)‖_∞ = tr(γ(γ₁ ⊗ δ₁)), where ‖γ₁‖₂ = ‖δ₁‖₂ = 1 and γ₁, δ₁ are Hermitian matrices.

Now, we have all the preliminary results required to discuss our new results.

3.

An Upper Bound for the Spectral Radius of the Special Triad

In this section we obtain an upper bound for the spectral radius of PPT states, SPC states and invariant under realignment states (Theorem 1). In order to prove this theorem, two lemmas are required.

Our first lemma says that the largest singular value of the partial transpose of any state cannot exceed the largest singular values of its reduced states or of its own realignment.

Lemma 4

Let γ ∈ ℳ_k ⊗ ℳ_k be any positive semidefinite Hermitian matrix. Then ${∥ γ^{Γ} ∥}_{\infty} \leq min {∥ γ_{A} ∥}_{\infty}, {∥ γ_{B} ∥}_{\infty}, {∥ R (γ) ∥}_{\infty}\} .$ {\left\| {\gamma ^\Gamma }\right\|_\infty } \le \min \left\{ {{\left\| {\gamma _A}\right\|_\infty },{\left\| {\gamma _B}\right\|_\infty },{\left\| {\cal R}(\gamma )\right\|_\infty }} \right\}.

Proof

Let v ∈ ℂ^k ⊗ ℂ^k be a unit vector such that |tr(γ^Γvv^*)| = ‖γ^Γ‖_∞. Let n be its rank. Denote by {g₁, …, g_k} and {e₁, …, e_n} the canonical bases of ℂ^k and ℂⁿ, respectively.

Next, there are matrices D ∈ ℳ_k×k, E ∈ ℳ_k×k, R ∈ ℳ_k×n and S ∈ ℳ_k×n such that

(1)
v = (D ⊗ Id)w, where tr(DD^*) = 1, $w = \sum_{i = 1}^{k} g_{i} \otimes g_{i} \in ℂ^{k} \otimes ℂ^{k}$ w=\sum\nolimits_{i=1}^{k}{}{{g}_{i}}\otimes {{g}_{i}}\in {{\mathbb{C}}^{k}}\otimes {{\mathbb{C}}^{k}} ,
(2)
v = (Id ⊗ E)w, where tr(E̅E^t) = 1, $w = \sum_{i = 1}^{k} g_{i} \otimes g_{i} \in ℂ^{k} \otimes ℂ^{k}$ w=\sum\nolimits_{i=1}^{k}{}{{g}_{i}}\otimes {{g}_{i}}\in {{\mathbb{C}}^{k}}\otimes {{\mathbb{C}}^{k}} ,
(3)
v = (R ⊗ S)u, where $u = \sum_{i = 1}^{n} e_{i} \otimes e_{i} \in ℂ^{n} \otimes ℂ^{n}$ u=\sum\nolimits_{i=1}^{n}{}{{e}_{i}}\otimes {{e}_{i}}\in {{\mathbb{C}}^{n}}\otimes {{\mathbb{C}}^{n}} and tr((RR∗)²) = tr((S̅S^t)²) = 1.

Now,

(1)
‖γ^Γ‖_∞ = |tr(γ^Γvv^*)| = |tr((D^* ⊗ Id)γ(D ⊗ Id)(ww^*)^Γ)| ≤ tr((D^* ⊗ Id)γ(D ⊗ Id)), since Id ⊗ Id ± (ww^*)^Γ and γ are positive semidefinite. Hence ${∥ γ^{Γ} ∥}_{\infty} \leq tr (γ ({DD}^{*} \otimes Id)) = tr (γ_{A} {DD}^{*}) \leq {∥ γ_{A} ∥}_{\infty} tr ({DD}^{*}) = {∥ γ_{A} ∥}_{\infty} .$ {\left\| {\gamma ^\Gamma }\right\|_\infty } \le tr(\gamma (D{D^*} \otimes Id)) = tr({\gamma _A}D{D^*}) \le {\left\| {\gamma _A}\right\|_\infty }tr(D{D^*}) = {\left\| {\gamma _A}\right\|_\infty }.
(2)
‖γ^Γ‖_∞ = |tr(γ^Γvv^*)| = |tr((Id ⊗ E^t)γ(Id ⊗ Ē)(ww^*)^Γ)| ≤ tr((Id ⊗ E^t)γ(Id ⊗ Ē)), since Id ⊗ Id ± (ww^*)^Γ and γ are positive semidefinite. Hence ${∥ γ^{Γ} ∥}_{\infty} \leq tr (γ (Id \otimes \bar{E} E^{t})) = tr (γ_{B} \bar{E} E^{t}) \leq {∥ γ_{B} ∥}_{\infty} tr (\bar{E} E^{t}) = {∥ γ_{B} ∥}_{\infty} .$ {\left\| {\gamma ^\Gamma }\right\|_\infty } \le tr(\gamma (Id \otimes \overline E {E^t})) = tr({\gamma _B}\overline E {E^t}) \le {\left\| {\gamma _B}\right\|_\infty }tr(\overline E {E^t}) = {\left\| {\gamma _B}\right\|_\infty }.
(3)
‖γ^Γ‖_∞ = |tr(γ^Γvv^*)| = |tr((R^* ⊗ S^t)γ(R ⊗ S̅)(uu^*)^Γ)| ≤ tr ((R^* ⊗ S^t)γ(R ⊗ S̅)), since Id ⊗ Id ± (uu^*)^Γ and γ are positive semidefinite. Hence ${∥ γ^{Γ} ∥}_{\infty} \leq tr (γ ({RR}^{*} \otimes \bar{S} S^{t})) \leq {∥ R (γ) ∥}_{\infty}, since {∥ {RR}^{*} ∥}_{2} = {∥ \bar{S} S^{t} ∥}_{2} = 1, by Lemma 3 .$ {\left\| {\gamma ^\Gamma }\right\|_\infty } \le tr(\gamma (R{R^*} \otimes \overline S {S^t})) \le {\left\| {\cal R}(\gamma )\right\|_\infty },{\rm{since}}{\left\| R{R^*}\right\|_2} = {\left\| \overline S {S^t}\right\|_2} = 1,\;{\rm{by}}\;{\rm{Lemma}}\;{\rm{3}}.

Our next lemma shows that the largest singular value of the realignment of any state cannot exceed the geometric mean of the largest singular values of its reduced states.

Lemma 5

Let γ ∈ ℳ_k ⊗ ℳ_k be a positive semidefinite Hermitian matrix. Then $∥ R (γ) ∥_{\infty}^{2} \leq {∥ γ_{A} ∥}_{\infty} {∥ γ_{B} ∥}_{\infty} .$ {\left\| {\cal R}(\gamma )\right\|_\infty ^2} \le {\left\| {\gamma _A}\right\|_\infty }{\left\| {\gamma _B}\right\|_\infty }.

Proof

By Lemma 3, there are Hermitian matrices γ₁ ∈ ℳ_k and δ₁ ∈ ℳ_k such that

(1)
$tr (γ_{1}^{2}) = tr (δ_{1}^{2}) = 1$ tr(\gamma _1^2) = tr(\delta _1^2) = 1
(2)
tr(γ(γ₁ ⊗ δ₁)) = ‖ℛ(γ)‖_∞.

Consider the following positive semidefinite Hermitian matrix $\begin{matrix} (\begin{matrix} γ^{\frac{1}{2}} & 0 \\ 0 & γ^{\frac{1}{2}} \end{matrix}) (\begin{matrix} Id \otimes δ_{1} \\ γ_{1} \otimes Id \end{matrix}) (\begin{matrix} Id \otimes δ_{1} & γ_{1} \otimes Id \end{matrix}) (\begin{matrix} γ^{\frac{1}{2}} & 0 \\ 0 & γ^{\frac{1}{2}} \end{matrix}) \\ = (\begin{matrix} γ^{\frac{1}{2}} (Id \otimes δ_{1}^{2}) γ^{\frac{1}{2}} & γ^{\frac{1}{2}} (γ_{1} \otimes δ_{1}) γ^{\frac{1}{2}} \\ γ^{\frac{1}{2}} (γ_{1} \otimes δ_{1}) γ^{\frac{1}{2}} & γ^{\frac{1}{2}} (γ_{1}^{2} \otimes Id) γ^{\frac{1}{2}} \end{matrix}) . \end{matrix}$ \matrix{ {\left( {\matrix{ {{\gamma ^{{1 \over 2}}}} & 0 \cr 0 & {{\gamma ^{{1 \over 2}}}} \cr } } \right)\left( {\matrix{ {Id \otimes {\delta _1}} \cr {{\gamma _1} \otimes Id} \cr } } \right)\left( {\matrix{ {Id \otimes {\delta _1}} & {{\gamma _1} \otimes Id} \cr } } \right)\left( {\matrix{ {{\gamma ^{{1 \over 2}}}} & 0 \cr 0 & {{\gamma ^{{1 \over 2}}}} \cr } } \right)} \cr { = \left( {\matrix{ {{\gamma ^{{1 \over 2}}}(Id \otimes \delta _1^2){\gamma ^{{1 \over 2}}}} & {{\gamma ^{{1 \over 2}}}({\gamma _1} \otimes {\delta _1}){\gamma ^{{1 \over 2}}}} \cr {{\gamma ^{{1 \over 2}}}({\gamma _1} \otimes {\delta _1}){\gamma ^{{1 \over 2}}}} & {{\gamma ^{{1 \over 2}}}(\gamma _1^2 \otimes Id){\gamma ^{{1 \over 2}}}} \cr } } \right).} \cr }

Its partial trace, $D = {(\begin{matrix} tr (γ ({Id}_{k} \otimes δ_{1}^{2})) & tr (γ (γ_{1} \otimes δ_{1})) \\ tr (γ (γ_{1} \otimes δ_{1})) & tr (γ (γ_{1}^{2} \otimes {Id}_{k})) \end{matrix})}_{2 \times 2}$ D = {\left( {\matrix{ {tr(\gamma (I{d_k} \otimes \delta _1^2))} & {tr(\gamma ({\gamma _1} \otimes {\delta _1}))} \cr {tr(\gamma ({\gamma _1} \otimes {\delta _1}))} & {tr(\gamma (\gamma _1^2 \otimes I{d_k}))} \cr } } \right)_{2 \times 2}} is also positive semidefinite.

Thus $0 \leq det (D) = tr (γ ({Id}_{k} \otimes δ_{1}^{2})) tr (γ (γ_{1}^{2} \otimes {Id}_{m})) - tr {(γ (γ_{1} \otimes δ_{1}))}^{2}$ 0 \le {\rm{det}}(D) = tr(\gamma (I{d_k} \otimes \delta _1^2))tr(\gamma (\gamma _1^2 \otimes I{d_m})) - tr{(\gamma ({\gamma _1} \otimes {\delta _1}))^2} .

Notice that

$tr (γ ({Id}_{k} \otimes δ_{1}^{2})) = tr (γ_{B} δ_{1}^{2}) \leq {∥ γ_{B} ∥}_{\infty} tr (δ_{1}^{2}) = {∥ γ_{B} ∥}_{\infty}$ tr(\gamma (I{d_k} \otimes \delta _1^2)) = tr({\gamma _B}\delta _1^2) \le {\left\| {\gamma _B}\right\|_\infty }tr(\delta _1^2) = {\left\| {\gamma _B}\right\|_\infty } ,
$tr (γ (γ_{1}^{2} \otimes {Id}_{m})) = tr (γ_{A} γ_{1}^{2}) \leq {∥ γ_{A} ∥}_{\infty} tr (γ_{1}^{2}) = {∥ γ_{A} ∥}_{\infty}$ tr(\gamma (\gamma _1^2 \otimes I{d_m})) = tr({\gamma _A}\gamma _1^2) \le {\left\| {\gamma _A}\right\|_\infty }tr(\gamma _1^2) = {\left\| {\gamma _A}\right\|_\infty } ,
$tr {(γ (γ_{1} \otimes δ_{1}))}^{2} = ∥ R (γ) ∥_{\infty}^{2}$ tr{(\gamma ({\gamma _1} \otimes {\delta _1}))^2} = {\left\| {\cal R}(\gamma )\right\|_\infty ^2} .

Hence $∥ R (γ) ∥_{\infty}^{2} \leq {∥ γ_{A} ∥}_{\infty} {∥ γ_{B} ∥}_{\infty}$ {\left\| {\cal R}(\gamma )\right\|_\infty ^2} \le {\left\| {\gamma _A}\right\|_\infty }{\left\| {\gamma _B}\right\|_\infty } .

These lemmas imply the first new connection for our special triad of quantum states. This theorem says that the spectral radius of any state that is PPT or SPC or invariant under realignment cannot exceed the largest singular values of its reduced states or of its own realignment.

Theorem 1

Let γ ∈ ℳ_k ⊗ ℳ_k be a positive semidefinite Hermitian matrix. If γ is PPT or SPC or invariant under realignment, then ${∥ γ ∥}_{\infty} \leq min {{∥ γ_{A} ∥}_{\infty}, {∥ γ_{B} ∥}_{\infty}, {∥ R (γ) ∥}_{\infty}} .$ {\left\| \gamma \right\|_\infty } \le \min \{ {\left\| {\gamma _A}\right\|_\infty },{\left\| {\gamma _B}\right\|_\infty },{\left\| {\cal R}(\gamma )\right\|_\infty }\} .

Proof

First, let γ be a PPT state. Hence γ^Γ is also a state.

Notice that (γ^Γ)_A = γ_A, ${(γ^{Γ})}_{B} = γ_{B}^{t}$ {({\gamma ^\Gamma })_B} = \gamma _B^t and, by lemma 3, ‖ℛ(γ^Γ)‖_∞ = ‖ℛ(γ)‖_∞.

By applying Lemma 4 on γ^Γ, we obtain ${∥ γ ∥}_{\infty} \leq min {{∥ {(γ^{Γ})}_{A} ∥}_{\infty}, {∥ {(γ^{Γ})}_{B} ∥}_{\infty}, {∥ R (γ^{Γ}) ∥}_{\infty}} = min {{∥ γ_{A} ∥}_{\infty}, {∥ γ_{B} ∥}_{\infty}, {∥ R (γ) ∥}_{\infty}} .$ {\left\| \gamma \right\|_\infty } \le \min \{ {\left\| {({\gamma ^\Gamma })_A}\right\|_\infty },{\left\| {({\gamma ^\Gamma })_B}\right\|_\infty },{\left\| {\cal R}({\gamma ^\Gamma })\right\|_\infty }\} = \min \{ {\left\| {\gamma _A}\right\|_\infty },{\left\| {\gamma _B}\right\|_\infty },{\left\| {\cal R}(\gamma )\right\|_\infty }\} .

So the proof of the PPT case is complete.

Next, if γ is SPC or invariant under realignment, then γ_A = γ_B or $γ_{A} = γ_{B}^{t}$ {\gamma _A} = \gamma _B^t by [5, Corollary 25]. Hence, by Lemma 5, ‖ℛ(γ)‖_∞ ≤ ‖γ_A‖_∞.

It remains to prove that ‖γ‖_∞ ≤ ‖ℛ(γ)‖_∞, whenever γ is SPC or invariant under realignment. Notice that this inequality is trivial for matrices invariant under realignment. Thus, let γ be an SPC state.

As defined in the introduction, ℛ(γ^Γ) is positive semidefinite. Applying Lemma 4 on ℛ(γ^Γ), we obtain ${∥ R {(γ^{Γ})}^{Γ} ∥}_{\infty} \leq {∥ R (R (γ^{Γ})) ∥}_{\infty} .$ {\left\| {\cal R}{({\gamma ^\Gamma })^\Gamma }\right\|_\infty } \le {\left\| {\cal R}({\cal R}({\gamma ^\Gamma }))\right\|_\infty }.

Now, by items (7) and (2) of Lemma 1, ℛ(γ^Γ)^Γ = γF and ℛ(ℛ(γ^Γ)) = γ^Γ, where F is the flip operator. Therefore ${∥ γ F ∥}_{\infty} \leq {∥ γ^{Γ} ∥}_{\infty} .$ {\left\| \gamma F\right\|_\infty } \le {\left\| {\gamma ^\Gamma }\right\|_\infty }.

Finally, ‖γ^Γ‖_∞ ≤ ‖ℛ(γ)‖_∞ by Lemma 4, and ‖γ‖_∞ = ‖γF ‖_∞, since F is unitary.

4.

Filter Normal Form for SPC States and Invariant under Realignment States

In this section we show that every SPC state and every invariant under realignment state can be put in the filter normal form. In addition their filter normal forms can still be chosen to be SPC and invariant under realignment, respectively (Theorem 2).

Then we show that every PPT state whose rank is equal to its two reduced ranks can be put in the filter normal form (see Theorem 3).

As described in the introduction, there are applications of this normal form in entanglement theory. Now, it has been noticed that this normal form is connected to an extension of Sinkhorn-Knopp theorem for positive maps [4,21]. This theorem concerns the existence of invertible matrices R, S such that R^*T (SXS^*)R is doubly stochastic for a positive map T (X) satisfying suitable conditions. So we start this section with some definitions and lemmas related to this theorem.

Let V ∈ ℳ_k be an orthogonal projection and consider the sub-algebra of ℳ_k : V ℳ_kV = {V XV, X ∈ ℳ_k}. Let P_k denote the set of positive semidefinite Hermitian matrices of ℳ_k.

Definition 2

Let us say that T : V ℳ_kV → V ℳ_kV is a positive map if T (X) ∈ P_k ∩ V ℳ_kV for every X ∈ P_k ∩ V ℳ_kV. In addition, we say that a positive map T : V ℳ_kV → V ℳ_kV is doubly stochastic if the following equivalent conditions hold:

(1)
the matrix A_m×m, defined as $A_{ij} = tr (T (v_{i} v_{i}^{*}) w_{j} w_{j}^{*})$ {A_{ij}} = tr(T({v_i}v_i^*){w_j}w_j^*) , is doubly stochastic for every choice of orthonormal bases v₁, …, v_m and w₁, …, w_m of Im(V),
(2)
T (V ) = T^*(V ) = V, where T^* : V ℳ_kV → V ℳ_kV is the adjoint of T : V ℳ_kV → V ℳ_kV with respect to the trace inner product.

Definition 3

A positive map T : V ℳ_kV → V ℳ_kV is said to be fully indecomposable if the following equivalent conditions hold:

(1)
the matrix A_m×m, defined as $A_{ij} = tr (T (v_{i} v_{i}^{*}) w_{j} w_{j}^{*})$ {A_{ij}} = tr(T({v_i}v_i^*){w_j}w_j^*) , is fully indecomposable [22] for every choice of orthonormal bases v₁, …, v_m and w₁, …, w_m of Im(V ),
(2)
rank(X) + rank(Y ) < rank(V ), whenever X, Y ∈ (V ℳ_kV ∩ P_k) ∖ {0}and tr(T (X)Y ) = 0,
(3)
rank(T (X)) > rank(X), ∀X ∈ V ℳ_kV ∩ P_k such that 0 < rank(X) < rank(V).

Below we prove two lemmas concerning self-adjoint maps with respect to the trace inner product.

Lemma 6

Let T : V ℳ_kV → V ℳ_kV be a fully indecomposable self-adjoint map. There is R ∈ V ℳ_kV such that R^*T (R(⋅)R^*)R : V ℳ_kV → V ℳ_kV is doubly stochastic.

Proof

Since T : V ℳ_kV → V ℳ_kV is fully indecomposable, it has total support [21, Lemma 2.3] or it has a positive achievable capacity [4]. So there are matrices A, B ∈ V ℳ_kV such that rank(A) = rank(B) = rank(V ) and T₁(X) = B^*T (AXA^*)B is doubly stochastic [21, Theorem 3.7]. Notice that T₁ still is fully indecomposable.

Now, $T_{2} (X) = T_{1}^{*} (X) = A^{*} T (B (\cdot) B^{*}) A$ {T_2}(X) = T_1^*(X) = {A^*}T(B( \cdot ){B^*})A is also doubly stochastic and (1) $T_{2} (X) = C^{*} T_{1} (DX D^{*}) C,$ {T_2}(X) = {C^*}{T_1}(DX{D^*})C, where C = B⁺A, D = A⁺B and Y⁺ is the pseudo-inverse of Y.

Let C = EF G^* and D = HLJ^* be the SVD decompositions of C and D, where

(1)
E = (e₁, …, e_m), G = (g₁, …, g_m), H = (h₁, …, h_m), J = (j₁, …, j_m) ∈ ℳ_k×m and the columns of each of these matrices form an orthonormal basis of Im(V).
(2)
F = diagonal(f₁, …, f_m), L = diagonal(l₁, …, l_m) and f_i > 0, l_i > 0 for every i.

Next, define R, S ∈ ℳ_m×m as $R_{ik} = tr (T_{2} (j_{i} j_{i}^{*}) g_{k} g_{k}^{*}) and S_{ik} = tr (T_{1} (h_{i} h_{i}^{*}) e_{k} e_{k}^{*}) .$ {R_{ik}} = tr({T_2}({j_i}j_i^*){g_k}g_k^*)\;{\rm{and}}\;{S_{ik}} = tr({T_1}({h_i}h_i^*){e_k}e_k^*).

By Equation (1), $R_{ik} = l_{i}^{2} f_{k}^{2} S_{ik}$ {R_{ik}} = l_i^2f_k^2\;{S_{ik}} , i.e., R = L² SF².

Thus, L², F² are positive diagonal matrices such that L²SF² is doubly stochastic by Definition 2. Recall that S is a fully indecomposable matrix by Definition 3.

Since S is fully indecomposable, by a theorem proved in [23], the diagonal matrices L² and F² such that L²SF² is doubly stochastic must be unique up to multiplication by positive numbers, but Id.S.Id is also doubly stochastic.

Thus, L = a⁻²Id and F = a²Id for some a > 0.

Therefore, B⁺A = C = a²U, where U = EG^*. Notice that U V U^* = V.

In addition, BB⁺A = a²BU. Since BB⁺ = V and V A = A, we obtain A = a²BU.

Thus, $B^{*} T (A (\cdot) A^{*}) B = B^{*} T ((a^{2} B) U (\cdot) U^{*} {(a^{2} B)}^{*}) B = {(aB)}^{*} T ((aB) U (\cdot) U^{*} {(aB)}^{*}) (aB) .$ {B^*}T(A( \cdot ){A^*})B = {B^*}T(({a^2}B)U( \cdot ){U^*}{({a^2}B)^*})B = {(aB)^*}T((aB)U( \cdot ){U^*}{(aB)^*})(aB).

Finally, (aB)^*T ((aB)(⋅)(aB)^*)(aB) is doubly stochastic too, since V = U V U^*.

Lemma 7

Let T : V ℳ_kV → V ℳ_kV be a self-adjoint positive map such that v ∉ ker(T (vv^*)) for every $v \in Im (V) \ {\vec{0}}$ v \in {\rm{Im}}(V)\backslash \{ \vec 0\} . Then there is R ∈ Vℳ_kV such that R^*T (R(⋅)R^*)R : Vℳ_kV → Vℳ_kV is doubly stochastic.

Proof

This proof is an induction on the rank(V).

If rank(V) = 1, then V ℳ_kV = {λvv^*, λ ∈ ℂ}. Thus, T (vv^*) = μvv^*, where μ > 0 by hypothesis.

Define $R = \frac{1}{\sqrt[4]{μ}} {vv}^{*}$ R = {1 \over {\root 4 \of \mu }}v{v^*} . So R^*T (Rvv^*R^*)R = vv^*. Thus, R^*T (R(⋅)R^*)R : V ℳ_kV → V ℳ_kV is a self-adjoint doubly stochastic map.

Let rank(V ) = n > 1 and assume the validity of this theorem whenever the rank of the orthogonal projection is less than n.

Consider all pairs of orthogonal projections (V₁, W₁) such that $V_{1}, W_{1} \in V M_{k} V \ {0} and 0 = tr (T (V_{1}) W_{1}) .$ {V_1},{W_1} \in V{{\cal M}_k}V\backslash \{ 0\} \;{\rm{and}}\;0 = tr(T({V_1}){W_1}).

Since there is no $v \in Im (V) \ {\vec{0}}$ v \in {\rm{Im}}(V)\backslash \{ \vec 0\} such that tr(T(vv^*)vv^*) = 0, $Im (V_{1}) \cap Im (W_{1}) = {\vec{0}}$ {\rm{Im}}({V_1}) \cap {\rm{Im}}({W_1}) = \{ \vec 0\} . So $rank (V_{1}) + rank (W_{1}) \leq rank (V) .$ {\rm{rank}}({V_1}) + {\rm{rank}}({W_1}) \le {\rm{rank}}(V).

If for every aforementioned pair (V₁, W₁), we have rank(V₁) + rank(W₁) < rank(V), then T is fully indecomposable by Definition 3. So the result follows by Lemma 6.

Let us assume that there is such a pair (V₁, W₁) satisfying rank(V₁) + rank(W₁) = rank(V).

Since $Im (V_{1}) \cap Im (W_{1}) = {\vec{0}}$ {\rm{Im}}({V_1}) \cap {\rm{Im}}({W_1}) = \{ \vec 0\} , there is S ∈ Vℳ_kV such that SV₁S^* = V₁ and S(V − V₁)S^* = W₁. Define T^′(X) = S^*T (SXS^*)S. Notice that tr(T^′(V₁)(V − V₁)) = 0.

Next, since T is self-adjoint so is T^′, hence tr(T^′(V − V₁)V₁) = 0.

These last two equalities imply that (2) $T^{'} (V_{1} M_{k} V_{1}) \subset V_{1} M_{k} V_{1} and T^{'} ((V - V_{1}) M_{k} (V - V_{1})) \subset (V - V_{1}) M_{k} (V - V_{1}) .$ T'({V_1}{{\cal M}_k}{V_1}) \subset {V_1}{{\cal M}_k}{V_1}\;\;{\rm{and}}\;\;T'((V - {V_1}){{\cal M}_k}(V - {V_1})) \subset (V - {V_1}){{\cal M}_k}(V - {V_1}).

Of course the restrictions T^′|_{V₁ℳ_kV₁} and T^′|_{(V−V₁)ℳ_k(V−V₁)} are self-adjoint and there is no $v \in Im (V_{1}) \ {\vec{0}}$ v \in {\mathop{\rm Im}\nolimits} ({V_1})\backslash \{ \vec 0\} or $v \in Im (V - V_{1}) \ {\vec{0}}$ v \in {\mathop{\rm Im}\nolimits} (V - {V_1})\backslash \{ \vec 0\} such that tr(T^′(vv^*)vv^*) = 0.

By induction hypothesis, there are R₁ ∈ V₁ℳ_kV₁ and R₂ ∈ (V − V₁)ℳ_k(V − V₁) such that

$R_{1}^{*} T^{'} (R_{1} (\cdot) R_{1}^{*}) R_{1} : V_{1} M_{k} V_{1} \to V_{1} M_{k} V_{1}$ R_1^*T'({R_1}( \cdot )R_1^*){R_1}:{V_1}{{\cal M}_k}{V_1} \to {V_1}{{\cal M}_k}{V_1} is doubly stochastic, i.e., (3) $R_{1}^{*} T^{'} (R_{1} (V_{1}) R_{1}^{*}) R_{1} = V_{1}$ R_1^*T'({R_1}({V_1})R_1^*){R_1} = {V_1}
$R_{2}^{*} T^{'} (R_{2} (\cdot) R_{2}^{*}) R_{2} : (V - V_{1}) M_{k} (V - V_{1}) \to (V - V_{1}) M_{k} (V - V_{1})$ R_2^*T'({R_2}( \cdot )R_2^*){R_2}:(V - {V_1}){{\cal M}_k}(V - {V_1}) \to (V - {V_1}){{\cal M}_k}(V - {V_1}) is doubly stochastic, i.e., (4) $R_{2}^{*} T^{'} (R_{2} (V - V_{1}) R_{2}^{*}) R_{2} = V - V_{1}$ R_2^*T'({R_2}(V - {V_1})R_2^*){R_2} = V - {V_1}

Set R = R₁ + R₂ ∈ V ℳ_kV. Notice that T^″(X) = R^*T^′(RXR^*)R is self-adjoint and $\begin{array}{l} T^{″} (V) & = T^{″} (V_{1} + V - V_{1}) = T^{″} (V_{1}) + T^{″} (V - V_{1}) \\ = R^{*} T^{'} ({RV}_{1} R^{*}) R + R^{*} T^{'} (R (V - V_{1}) R^{*}) R \\ = R^{*} T^{'} (R_{1} V_{1} R_{1}^{*}) R + R^{*} T^{'} (R_{2} (V - V_{1}) R_{2}^{*}) R \\ = R_{1}^{*} T^{'} (R_{1} V_{1} R_{1}^{*}) R_{1} + R_{2}^{*} T^{'} (R_{2} (V - V_{1}) R_{2}^{*}) R_{2}, by equation 2, \\ = V_{1} + (V - V_{1}) = V, by equations 3 and 4. \end{array}$ \matrix{ {T''(V)} \hfill & { = T''({V_1} + V - {V_1}) = T''({V_1}) + T''(V - {V_1})} \hfill \cr {} \hfill & { = {R^*}T'(R{V_1}{R^*})R + {R^*}T'(R(V - {V_1}){R^*})R} \hfill \cr {} \hfill & { = {R^*}T'({R_1}{V_1}R_1^*)R + {R^*}T'({R_2}(V - {V_1})R_2^*)R} \hfill \cr {} \hfill & { = R_1^*T'({R_1}{V_1}R_1^*){R_1} + R_2^*T'({R_2}(V - {V_1})R_2^*){R_2},\;{\rm{by}}\;{\rm{equation}}\;2,} \hfill \cr {} \hfill & { = {V_1} + (V - {V_1}) = V,\;{\rm{by}}\;{\rm{equations}}\;3\;{\rm{and}}\;4.} \hfill \cr }

Hence, T^″ : V ℳ_kV → V ℳ_kV is doubly stochastic.

Corollary 1

Let A ∈ ℳ_k⊗ℳ_k be an Hermitian matrix such that G_A : ℳ_k → ℳ_k is a self-adjoint positive map and tr(A(vv^* ⊗ vv^*)) > 0 for every v ∈ ℂ^k ∖{0}. There is an invertible matrix R ∈ ℳ_k such that $(R^{*} \otimes R^{*}) A (R \otimes R) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i}$ ({R^*} \otimes {R^*})A(R \otimes R) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i} , where

(1)
λ₁ = 1 and $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }}
(2)
λ_i ∈ ℝ and $γ_{i} = γ_{i}^{*}$ {\gamma _i} = \gamma _i^* for every i,
(3)
1 ≥ |λ_i| for every i,
(4)
tr(γ_iγ_j) = 0 for every i ≠ j and $tr (γ_{i}^{2}) = 1$ tr(\gamma _i^2) = 1 for every i.

Proof

By the definition of G_A : ℳ_k → ℳ_k (given in the introduction), notice that $0 < tr (A ({vv}^{*} \otimes {vv}^{*})) = tr (G_{A} ({vv}^{*}) {vv}^{*}) .$ 0 < tr(A(v{v^*} \otimes v{v^*})) = tr({G_A}(v{v^*})v{v^*}).

Hence v ∉ ker G_A(vv^*) for every v ∈ ℂ^k ∖ {0}.

By Lemma 7, there is an invertible matrix R ∈ ℳ_k such that R^*G_A(RXR^*)R is doubly stochastic.

Define B = (R^* ⊗ R^*)A(R ⊗ R) and notice that G_B(X) = R^*G_A(RXR^*)R. Therefore, G_B is a self-adjoint doubly stochastic map, i.e., $G_{B} (\frac{Id}{\sqrt{k}}) = \frac{Id}{\sqrt{k}}$ {G_B}\left( {{{Id} \over {\sqrt k }}} \right) = {{Id} \over {\sqrt k }} .

Let $γ_{1} = \frac{Id}{\sqrt{k}}, γ_{2}, \dots, γ_{k^{2}}$ {\gamma _1} = {{Id} \over {\sqrt k }},{\gamma _2}, \ldots ,{\gamma _{{k^2}}} be an orthonormal basis of ℳ_k formed by Hermitian eigenvectors of the self-adjoint positive map G_B : ℳ_k → ℳ_k such that

G_B(γ_i) = λ_iγ_i, where |λ_i| > 0 for 1 ≤ i ≤ n
G_B(γ_i) = 0, for i > n.

Since G_B is a positive map satisfying G_B(Id) = Id, its spectral radius is 1 [24, Theorem 2.3.7]. So |λ_i| ≤ 1 for every i.

Finally, by the definition of G_B, $B = \frac{Id}{\sqrt{k}} \otimes G_{B} (\frac{Id}{\sqrt{k}}) + γ_{2} \otimes G_{B} (γ_{2}) + \dots + γ_{k^{2}} \otimes G_{B} (γ_{k^{2}}) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i} .$ B = {{Id} \over {\sqrt k }} \otimes {G_B}\left( {{{Id} \over {\sqrt k }}} \right) + {\gamma _2} \otimes {G_B}({\gamma _2}) + \ldots + {\gamma _{{k^2}}} \otimes {G_B}({\gamma _{{k^2}}}) = \sum\limits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i}.

The next theorem shows that SPC states and invariant under realignment states can be put in the filter normal form retaining their SPC structure and invariance under realignment, respectively.

Theorem 2

Let γ ∈ ℳ_k ⊗ ℳ_k be a positive semidefinite Hermitian matrix such that rank(γ_A) = k. There is an invertible matrix R ∈ ℳ_k such that

(1)
$(R^{*} \otimes R^{*}) γ (R \otimes R) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i}$ ({R^*} \otimes {R^*})\gamma (R \otimes R) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i} , if ℛ(γ^Γ) is positive semidefinite;
(2)
$(R^{*} \otimes R^{t}) γ (R \otimes \bar{R}) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes \bar{γ_{i}}$ ({R^*} \otimes {R^t})\gamma (R \otimes \overline R ) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes \overline {{\gamma _i}} , if ℛ(γ) is positive semidefinite,

where
- a)
  $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} and $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }}
- b)
  $\frac{1}{k} \geq λ_{i} > 0$ {1 \over k} \ge {\lambda _i} > 0 and $γ_{i} = γ_{i}^{*}$ {\gamma _i} = \gamma _i^* for every i,
- c)
  tr(γ_iγ_j) = 0 for every i ≠ j and $tr (γ_{i}^{2}) = 1$ tr(\gamma _i^2) = 1 for every i.

Proof

(1) If γ is a state such that ℛ(γ^Γ) is positive semidefinite, then, by [5, corollary 25], γ can be written as $γ = \sum_{i = 1}^{n} a_{i} B_{i} \otimes B_{i}$ \gamma = \sum\nolimits_{i = 1}^n {a_i}{B_i} \otimes {B_i} , where a_i > 0, $B_{i} = B_{i}^{*}$ {B_i} = B_i^* and $tr (B_{i}^{2}) = 1$ tr(B_i^2) = 1 for every i, and tr(B_iB_j) = 0 for i ≠ j.

Hence $G_{γ} (X) = \sum_{i = 1}^{n} a_{i} B_{i} tr (B_{i} X)$ {G_\gamma }(X) = \sum\nolimits_{i = 1}^n {a_i}{B_i}tr({B_i}X) is a self-adjoint map with positive eigenvalues a₁, …, a_n and possibly some null eigenvalues. In addition, since γ is positive semidefinite, G_γ (X) is a positive map.

Now, let v ∈ ℂ^k be such that $0 = tr (γ ({vv}^{*} \otimes {vv}^{*})) = \sum_{i = 1}^{n} a_{i} tr {(B_{i} {vv}^{*})}^{2} .$ 0 = tr(\gamma (v{v^*} \otimes v{v^*})) = \sum\nolimits_{i = 1}^n {a_i}tr{({B_i}v{v^*})^2}.

Since a_i > 0 and tr(B_i vv^*) ∈ ℝ for every i, tr(B_i vv^*) = 0 for every i. Therefore, $tr (γ_{A} {vv}^{*}) = \sum_{i = 1}^{n} a_{i} tr (B_{i}) tr (B_{i} {vv}^{*}) = 0 .$ tr({\gamma _A}v{v^*}) = \sum\nolimits_{i = 1}^n {a_i}tr({B_i})tr({B_i}v{v^*}) = 0.

By hypothesis γ_A is positive definite, hence v = 0.

So, by Corollary 1, there is a invertible matrix R such that (5) $(R^{*} \otimes R^{*}) γ (R \otimes R) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i}$ ({R^*} \otimes {R^*})\gamma (R \otimes R) = \sum\limits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i} satisfies the four conditions of that corollary. It remains to show that λ_i > 0 and then we multiply Equation (5) by $\frac{1}{k}$ {1 \over k} to obtain our desired result.

Finally, since G_γ has only non-negative eigenvalues and λ₁, …, λ_n are non-null eigenvalues of R^*G_γ (RXR^*)R (as seen in the proof of Corollary 1), λ₁, …, λ_n are positive.

(2) If γ is a state such that ℛ(γ) is positive semidefinite, then, by [5, Corollary 25], γ can be written as $γ = \sum_{i = 1}^{n} a_{i} B_{i} \otimes \bar{B_{i}}$ \gamma = \sum\nolimits_{i = 1}^n {a_i}{B_i} \otimes \overline {{B_i}} , where a_i > 0, $B_{i} = B_{i}^{*}$ {B_i} = B_i^* and $tr (B_{i}^{2}) = 1$ tr(B_i^2) = 1 for every i, and tr(B_iB_j) = 0 for i ≠ j.

Consider $γ^{Γ} = \sum_{i = 1}^{n} a_{i} B_{i} \otimes B_{i}$ {\gamma ^\Gamma } = \sum\nolimits_{i = 1}^n {a_i}{B_i} \otimes {B_i} and notice that G_γ^Γ(X) = G_γ (X)^t is also a positive map. Now repeat the proof of item (1) for γ^Γ. Hence there is a invertible matrix R such that (6) $(R^{*} \otimes R^{*}) γ^{Γ} (R \otimes R) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i},$ ({R^*} \otimes {R^*}){\gamma ^\Gamma }(R \otimes R) = \sum\limits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i}, where γ_i and λ_i satisfy all the required conditions. Finally, $(R^{*} \otimes R^{t}) γ (R \otimes \bar{R}) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes \bar{γ_{i}} .$ ({R^*} \otimes {R^t})\gamma (R \otimes \overline R ) = \sum\limits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes \overline {{\gamma _i}} .

The next corollary says that after a local operation any state possess a Schmidt decomposition such that one of its matrices is a multiple of the identity.

Corollary 2

Let γ ∈ ℳ_k ⊗ ℳ_k be a state such that rank(γ_A) = k. There is an invertible matrix R ∈ ℳ_k such that $(R^{*} \otimes Id) γ (R \otimes Id) = \sum_{i = 1}^{n} a_{i} γ_{i} \otimes δ_{i}$ , where

a)
a₁ ≥ a_i > 0, for every 1 ≤ i ≤ n, and $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }} ,
b)
$γ_{i} = γ_{i}^{*}$ {\gamma _i} = \gamma _i^* , $δ_{i} = δ_{i}^{*}$ {\delta _i} = \delta _i^* for every i,
c)
tr(γ_iγ_j) = tr(δ_iδ_j) = 0 for every i ≠ j and $tr (γ_{i}^{2}) = tr (δ_{i}^{2}) = 1$ tr(\gamma _i^2) = tr(\delta _i^2) = 1 .

Proof

First, since γ is a state, so is $F \bar{γ} F$ F\overline \gamma F . Therefore, $γ * F \bar{γ} F$ \gamma *F\overline \gamma F is positive semidefinite by item a) of Remark 1.

Now, by items (8) and (9) of Lemma 1, $R (γ * F \bar{γ} F) = R (γ) R {(γ)}^{*} .$ {\cal R}(\gamma *F\overline \gamma F) = {\cal R}(\gamma ){\cal R}{(\gamma )^*}.

It is not difficult to check that $rank ({(γ * F \bar{γ} F)}_{A}) = k$ {\rm{rank}}({(\gamma *F\overline \gamma F)_A}) = k , we leave it for last.

By item (2) of Theorem 2, there is an invertible matrix R ∈ ℳ_k such that $(R^{*} \otimes R^{t}) (γ * F \bar{γ} F) (R \otimes \bar{R}) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i},$ ({R^*} \otimes {R^t})(\gamma *F\overline \gamma F)(R \otimes \overline R ) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i}, where

a)
$λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} and $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }} ,
b)
$\frac{1}{k} \geq λ_{i} > 0$ {1 \over k} \ge {\lambda _i} > 0 and $γ_{i} = γ_{i}^{*}$ {\gamma _i} = \gamma _i^* for every i,
c)
tr(γ_iγ_j) = 0 for every i ≠ j and $tr (γ_{i}^{2}) = 1$ tr(\gamma _i^2) = 1 for every i.

Define δ = (R^* ⊗ Id)γ(R ⊗ Id) and notice that $δ * F δ^{t} F = δ * F \bar{δ} F = (R^{*} \otimes R^{t}) (γ * F \bar{γ} F) (R \otimes \bar{R}) .$ \delta *F{\delta ^t}F = \delta *F\overline \delta F = ({R^*} \otimes {R^t})(\gamma *F\overline \gamma F)(R \otimes \overline R ).

Thus, $G_{δ * F δ^{t} F} (\frac{Id}{k}) = λ_{1} \frac{Id}{k}$ {G_{\delta *F{\delta ^t}F}}\left( {{{Id} \over k}} \right) = {\lambda _1}{{Id} \over k} .

By item c) of Remark 1, $F_{δ} (G_{δ} (\frac{Id}{\sqrt{k}})) = G_{δ * F δ^{t} F} {(\frac{Id}{\sqrt{k}})}^{t} = λ_{1} \frac{Id}{\sqrt{k}}$ {F_\delta }\left( {{G_\delta }\left( {{{Id} \over {\sqrt k }}} \right)} \right) = {G_{\delta *F{\delta ^t}F}}{\left( {{{Id} \over {\sqrt k }}} \right)^t} = {\lambda _1}{{Id} \over {\sqrt k }} . So F_δ(G_δ(Id)) = λ₁Id.

By [24, Theorem 2.3.7], λ₁ is the largest eigenvalue of the positive map F_δ ∘ G_δ. So $\sqrt{λ_{1}}$ \sqrt {{\lambda _1}} is the largest singular value of G_δ and F_δ, since they are adjoints.

Next, let $δ_{1} = \frac{Id}{\sqrt{k}}, δ_{2}, \dots, δ_{k^{2}}$ {\delta _1} = {{Id} \over {\sqrt k }},{\delta _2}, \ldots ,{\delta _{{k^2}}} be an orthonormal basis of ℳ_k formed by Hermitian eigenvectors of F_δ ∘ G_δ : ℳ_k → ℳ_k.

Notice that $(R^{*} \otimes Id) γ (R \otimes Id) = δ = \sum_{i = 1}^{k^{2}} δ_{i} \otimes G_{δ} (δ_{i})$ ({R^*} \otimes Id)\gamma (R \otimes Id) = \delta = \sum\nolimits_{i = 1}^{{k^2}} {\delta _i} \otimes {G_\delta }({\delta _i}) .

If G_δ(δ_i) ≠ 0_k×k, for 1 ≤ i ≤ n, then define a_i = ‖G_δ(δ_i)‖₂ > 0. Thus, $δ = \sum_{i = 1}^{n} a_{i} δ_{i} \otimes \frac{1}{a_{i}} G_{δ} (δ_{i})$ \delta = \sum\nolimits_{i = 1}^n {a_i}{\delta _i} \otimes {1 \over {{a_i}}}{G_\delta }({\delta _i}) .

Notice that $G_{δ} {(δ_{i})}^{*} = G_{δ^{*}} (δ_{i}^{*}) = G_{δ} (δ_{i})$ {G_\delta }{({\delta _i})^*} = {G_{{\delta ^*}}}(\delta _i^*) = {G_\delta }({\delta _i}) , since δ and δ_i are Hermitian matrices. Moreover, by the definition of a_i, $tr (\frac{1}{a_{i}} G_{δ} (δ_{i}) \frac{1}{a_{i}} G_{δ} (δ_{i})) = 1, 1 \leq i \leq n .$ tr\left( {{1 \over {{a_i}}}{G_\delta }({\delta _i}){1 \over {{a_i}}}{G_\delta }({\delta _i})} \right) = 1,\;\;\;1 \le i \le n.

In addition, since F_δ(G_δ(δ_j))) is a multiple of δ_j, δ_i is orthogonal to F_δ(G_δ(δ_j))) for i ≠ j, $tr (\frac{1}{a_{i}} G_{δ} (δ_{i}) \frac{1}{a_{j}} G_{δ} (δ_{j})) = \frac{1}{a_{i} a_{j}} tr (δ_{i} F_{δ} (G_{δ} (δ_{j}))) = 0 .$ tr\left( {{1 \over {{a_i}}}{G_\delta }({\delta _i}){1 \over {{a_j}}}{G_\delta }({\delta _j})} \right) = {1 \over {{a_i}{a_j}}}tr({\delta _i}{F_\delta }({G_\delta }({\delta _j}))) = 0.

Finally, $a_{1}^{2} = tr (G_{δ} {(δ_{1})}^{2}) = tr (δ_{1} F_{δ} (G_{δ} (δ_{1}))) = λ_{1} tr (δ_{1}^{2}) = λ_{1}$ a_1^2 = tr({G_\delta }{({\delta _1})^2}) = tr({\delta _1}{F_\delta }({G_\delta }({\delta _1}))) = {\lambda _1}tr(\delta _1^2) = {\lambda _1} , so $a_{1} = \sqrt{λ_{1}}$ {a_1} = \sqrt {{\lambda _1}} . Notice also that, for every i, a_i ≤ the largest singular value of G_δ, which is $\sqrt{λ_{1}} = a_{1}$ \sqrt {{\lambda _1}} = {a_1} .

It remains to check that $rank ({(γ * F \bar{γ} F)}_{A}) = k$ {\rm{rank}}({(\gamma *F\overline \gamma F)_A}) = k . Indeed, since F_γ and G_γ are adjoints, Im(F_γ ∘ G_γ ) = Im(F_γ ). So there are Hermitian matrices Y, Z ∈ ℳ_k such that F_γ (G_γ (Y + iZ)) = F_γ (Id) = γ_A, which is positive definite. Since F_γ and G_γ are also positive maps, they leave the set of Hermitian matrices invariant, hence F_γ (G_γ (Y )) = γ_A.

There is λ > 0 such that Id − λY is positive semidefinite, so F_γ (G_γ (Id − λY )) is positive semidefinite too. Thus, F_γ (G_γ (Id)) = F_γ (G_γ (Id − λY )) + λγ_A, which is positive definite.

Finally, since γ and $γ * F \bar{γ} F$ \gamma *F\overline \gamma F are Hermitian matrices $F_{γ * F \bar{γ} F} (\cdot) = G_{γ * F \bar{γ} F}^{*} (\cdot) = {({(F_{γ} \circ G_{γ})}^{t})}^{*} = G_{γ}^{*} \circ F_{γ}^{*} ({(\cdot)}^{t}) = F_{γ} \circ G_{γ} ({(\cdot)}^{t}),$ {F_{\gamma *F\overline \gamma F}}( \cdot ) = G_{\gamma *F\overline \gamma F}^*( \cdot ) = {({({F_\gamma } \circ {G_\gamma })^t})^*} = G_\gamma ^* \circ F_\gamma ^*({( \cdot )^t}) = {F_\gamma } \circ {G_\gamma }({( \cdot )^t}), where the second equality comes from item c) of Remark 1.

Therefore ${(γ * F \bar{γ} F)}_{A} = F_{γ * F \bar{γ} F} (Id) = F_{γ} (G_{γ} (Id))$ {(\gamma *F\overline \gamma F)_A} = {F_{\gamma *F\overline \gamma F}}(Id) = {F_\gamma }({G_\gamma }(Id)) , which is positive definite.

For the last result of this section, we finally prove that every PPT state whose rank coincides with its two reduced ranks can be put in the filter normal form. This is the key result to prove Theorem 6. See how the relations between linear contractions (Proposition 1) are important in proving this result.

Theorem 3

If γ ∈ ℳ_k ⊗ ℳ_k is a PPT state such that rank(γ) = rank(γ_A) = rank(γ_B) = k, then there are invertible matrices R, S ∈ ℳ_k such that $δ_{A} = δ_{B} = \frac{1}{k} Id$ {\delta _A} = {\delta _B} = {1 \over k}Id , where δ = (R ⊗ S)γ(R^* ⊗ S^*).

Proof

First, define γ₁ = (Id ⊗ S)γ(Id ⊗ S^*) such that ${(γ_{1})}_{B} = \frac{1}{k} Id$ {({\gamma _1})_B} = {1 \over k}Id , where S is invertible.

Since γ₁ is also PPT, ${∥ γ_{1} ∥}_{\infty} \leq {∥ {(γ_{1})}_{B} ∥}_{\infty} = \frac{1}{k}$ {\left\| {\gamma _1}\right\|_\infty } \le {\left\| {({\gamma _1})_B}\right\|_\infty } = {1 \over k} , by Theorem 1.

Hence $1 = tr ({(γ_{1})}_{B}) = tr (γ_{1}) \leq {∥ γ_{1} ∥}_{\infty} rank (γ_{1}) = \frac{1}{k} . k = 1$ 1 = tr({({\gamma _1})_B}) = tr({\gamma _1}) \le {\left\| {\gamma _1}\right\|_\infty }\,{\rm{rank}}({\gamma _1}) = {1 \over k}.k = 1 . So γ₁ has k eigenvalues equal to $\frac{1}{k}$ {1 \over k} and the others 0.

Notice that ${(F \bar{γ_{1}} F)}^{Γ}$ {(F\overline {{\gamma _1}} F)^\Gamma } is positive semidefinite and so is ${(γ_{1} * F \bar{γ_{1}} F)}^{Γ} = γ_{1} * {(F \bar{γ_{1}} F)}^{Γ}$ {({\gamma _1}*F\overline {{\gamma _1}} F)^\Gamma } = {\gamma _1}*{(F\overline {{\gamma _1}} F)^\Gamma } as a *–product of two positive semidefinite Hermitian matrices by item a) of Remark 1. So the positive semidefinite Hermitian matrix $γ_{1} * F \bar{γ_{1}} F$ {\gamma _1}*F\overline {{\gamma _1}} F is PPT.

Now, by items (8) and (9) of Lemma 1, $R (γ_{1} * F \bar{γ_{1}} F) = R (γ_{1}) R {(γ_{1})}^{*}$ {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F) = {\cal R}({\gamma _1}){\cal R}{({\gamma _1})^*} , which is positive semidefinite.

Next, on the one hand $tr (R (γ_{1} * F \bar{γ_{1}} F)) = tr (R (γ_{1}) R {(γ_{1})}^{*}) = tr (γ_{1} γ_{1}^{*}) = \frac{1}{k}$ tr({\cal R}({\gamma _1}*F\overline {{\gamma _1}} F)) = tr({\cal R}({\gamma _1}){\cal R}{({\gamma _1})^*}) = tr({\gamma _1}\gamma _1^*) = {1 \over k} , since ℛ is an isometry.

On the other hand $\begin{array}{l} {R {(γ_{1} * F \bar{γ_{1}} F)}^{Γ}‖}_{1} & = {R (γ_{1} * (F \bar{γ_{1}} F) F)‖}_{1} (by item (6) of Lemma 1) \\ = {∥ R (γ_{1} * (F \bar{γ_{1}} F) F) F ∥}_{1} (since F is unitary) \\ = {∥ {(γ_{1} * (F \bar{γ_{1}} F))}^{Γ} ∥}_{1} (by item (4) of Lemma 1) \\ = tr (γ_{1} * (F \bar{γ_{1}} F)) (since γ_{1} * (F \bar{γ_{1}} F) is PPT) \\ = tr ({(γ_{1})}_{B} {(γ_{1})}_{B}^{*}) = \frac{1}{k} (by item c) of Remark 1) . \end{array}$ \matrix{ {{{\left\| {{\cal R}{{({\gamma _1}*F\overline {{\gamma _1}} F)}^\Gamma }} \right\|}_1}} \hfill & { = \;{{\left\| {{\cal R}({\gamma _1}*(F\overline {{\gamma _1}} F)F)} \right\|}_1}\;({\rm{by}}\;{\rm{item}}\;\left( 6 \right)\;{\rm{of}}\;{\rm{Lemma}}\;1)} \hfill \cr {} \hfill & { = \;{\left\| {\cal R}({\gamma _1}*(F\overline {{\gamma _1}} F)F)F\right\|_1}\;({\rm{since}}\;F\;{\rm{is}}\;{\rm{unitary}})} \hfill \cr {} \hfill & { = \;{\left\| {{({\gamma _1}*(F\overline {{\gamma _1}} F))}^\Gamma }\right\|_1}\;({\rm{by}}\;{\rm{item}}\;\left( 4 \right)\;{\rm{of}}\;{\rm{Lemma}}\;1)} \hfill \cr {} \hfill & { = tr({\gamma _1}*(F\overline {{\gamma _1}} F))\;({\rm{since}}\;{\gamma _1}*(F\overline {{\gamma _1}} F)\;{\rm{is}}\;{\rm{PPT}})} \hfill \cr {} \hfill & { = tr({{({\gamma _1})}_B}({\gamma _1})_B^*) = {1 \over k}\;({\rm{by}}\;{\rm{item}}\;c)\;{\rm{of}}\;{\rm{Remark}}\;1).} \hfill \cr }

Therefore, $tr (R (γ_{1} * F \bar{γ_{1}} F)) = {∥ R {(γ_{1} * F \bar{γ_{1}} F)}^{Γ} ∥}_{1}$ tr({\cal R}({\gamma _1}*F\overline {{\gamma _1}} F)) = {\left\| {\cal R}{({\gamma _1}*F\overline {{\gamma _1}} F)^\Gamma }\right\|_1} .

Since $R (γ_{1} * F \bar{γ_{1}} F)$ {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F) is positive semidefinite, this last equality means that $R {(γ_{1} * F \bar{γ_{1}} F)}^{Γ}$ {\cal R}{({\gamma _1}*F\overline {{\gamma _1}} F)^\Gamma } is positive semidefinite, i.e., $R (γ_{1} * F \bar{γ_{1}} F)$ {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F) is PPT.

We have just discovered that $R (γ_{1} * F \bar{γ_{1}} F)$ {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F) and $γ_{1} * F \bar{γ_{1}} F$ {\gamma _1}*F\overline {{\gamma _1}} F are PPT, but in this situation Lemma 2 says that $γ_{1} * F \bar{γ_{1}} F = R (γ_{1} * F \bar{γ_{1}} F)$ {\gamma _1}*F\overline {{\gamma _1}} F = {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F) .

Now, by Corollary 2, there is an invertible matrix R ∈ ℳ_k such that $γ_{2} = (R^{*} \otimes Id) γ_{1} (R \otimes Id) = \sum_{i = 1}^{n} λ_{i} A_{i} \otimes B_{i},$ {\gamma _2} = ({R^*} \otimes Id){\gamma _1}(R \otimes Id) = \sum\limits_{i = 1}^n {\lambda _i}{A_i} \otimes {B_i}, where

a)
λ₁ ≥ λ_i > 0 for every i and $A_{1} = \frac{Id}{\sqrt{k}}$ {A_1} = {{Id} \over {\sqrt k }} ,
b)
$A_{i} = A_{i}^{*}$ {A_i} = A_i^* , $B_{i} = B_{i}^{*}$ {B_i} = B_i^* for every i,
c)
tr(A_iA_j) = tr(B_iB_j) = 0 for every i ≠ j and $tr (A_{i}^{2}) = tr (B_{i}^{2}) = 1$ tr(A_i^2) = tr(B_i^2) = 1 .

In addition, we can normalize its trace, so assume that tr(γ₂) = 1.

Hence, $γ_{2} * (F \bar{γ_{2}} F) = \sum_{i = 1}^{n} λ_{i}^{2} A_{i} \otimes \bar{A_{i}} = (R^{*} \otimes R^{t}) (γ_{1} * (F \bar{γ_{1}} F)) (R \otimes \bar{R})$ {\gamma _2}*(F\overline {{\gamma _2}} F) = \sum\nolimits_{i = 1}^n \lambda _i^2{A_i} \otimes \overline {{A_i}} = ({R^*} \otimes {R^t})({\gamma _1}*(F\overline {{\gamma _1}} F))(R \otimes \overline R ) .

As $γ_{1} * (F \bar{γ_{1}} F)$ {\gamma _1}*(F\overline {{\gamma _1}} F) , the positive semidefinite Hermitian matrix $γ_{2} * (F \bar{γ_{2}} F)$ {\gamma _2}*(F\overline {{\gamma _2}} F) is also invariant under realignment because $\begin{array}{l} R (γ_{2} * (F \bar{γ_{2}} F)) & = R ((R^{*} \otimes R^{t}) (γ_{1} * (F \bar{γ_{1}} F)) (R \otimes \bar{R})) \\ = (R^{*} \otimes R^{t}) R (γ_{1} * (F \bar{γ_{1}} F)) (R \otimes \bar{R}), by item (3) of lemma 1, \\ = (R^{*} \otimes R^{t}) (γ_{1} * (F \bar{γ_{1}} F)) (R \otimes \bar{R}), since γ_{1} * F \bar{γ_{1}} F = R (γ_{1} * F \bar{γ_{1}} F) . \\ = γ_{2} * (F \bar{γ_{2}} F) . \end{array}$ \matrix{ {{\cal R}({\gamma _2}*(F\overline {{\gamma _2}} F))} \hfill & { = {\cal R}(({R^*} \otimes {R^t})({\gamma _1}*(F\overline {{\gamma _1}} F))(R \otimes \overline R ))} \hfill \cr {} \hfill & { = ({R^*} \otimes {R^t}){\cal R}({\gamma _1}*(F\overline {{\gamma _1}} F))(R \otimes \overline R ),\;{\rm{by}}\;{\rm{item}}\;(3)\;{\rm{of}}\;{\rm{lemma}}\;1,} \hfill \cr {} \hfill & { = ({R^*} \otimes {R^t})({\gamma _1}*(F\overline {{\gamma _1}} F))(R \otimes \overline R ),\;{\rm{since}}\;{\gamma _1}*F\overline {{\gamma _1}} F = {\cal R}({\gamma _1}*F\overline {{\gamma _1}} F).} \hfill \cr {} \hfill & { = {\gamma _2}*(F\overline {{\gamma _2}} F).} \hfill \cr } $\begin{array}{l} Now, notice that k λ_{1}^{2} & = tr (γ_{2} * (F \bar{γ_{2}} F) (A_{1} \otimes \bar{A_{1}})) k \\ = tr (γ_{2} * (F \bar{γ_{2}} F) (\frac{Id}{\sqrt{k}} \otimes \frac{Id}{\sqrt{k}})) k \\ = tr (γ_{2} * (F \bar{γ_{2}} F)) \\ = tr (R (γ_{2} * (F \bar{γ_{2}} F))), since γ_{2} * F \bar{γ_{2}} F = R (γ_{2} * F \bar{γ_{2}} F) \\ = tr (R (γ_{2}) R {(γ_{2})}^{*}), by item (8 - 9) of Lemma 1 \\ = tr (γ_{2} γ_{2}^{*}), since R is an isometry . \end{array}$ \matrix{ {{\rm{Now}},\;{\rm{notice}}\;{\rm{that}}\;k\lambda _1^2} \hfill & { = tr\left( {{\gamma _2}*(F\overline {{\gamma _2}} F)\;({A_1} \otimes \overline {{A_1}} )} \right)k} \hfill \cr {} \hfill & { = tr\left( {{\gamma _2}*(F\overline {{\gamma _2}} F)\;({{Id} \over {\sqrt k }} \otimes {{Id} \over {\sqrt k }})} \right)k} \hfill \cr {} \hfill & { = tr({\gamma _2}*(F\overline {{\gamma _2}} F))} \hfill \cr {} \hfill & { = tr({\cal R}({\gamma _2}*(F\overline {{\gamma _2}} F))),\;{\rm{since}}\;{\gamma _2}*F\overline {{\gamma _2}} F = {\cal R}({\gamma _2}*F\overline {{\gamma _2}} F)} \hfill \cr {} \hfill & { = \;tr({\cal R}({\gamma _2}){\cal R}{{({\gamma _2})}^*}),\;{\rm{by}}\;{\rm{item}}\;(8 - 9)\;{\rm{of}}\;{\rm{Lemma}}\;1} \hfill \cr {} \hfill & { = tr({\gamma _2}\gamma _2^*),\;{\rm{since}}\;{\cal R}\;{\rm{is}}\;{\rm{an}}\;{\rm{isometry}}.} \hfill \cr }

Next, ${∥ γ_{2} ∥}_{\infty}^{2} \leq {∥ R (γ_{2}) ∥}_{\infty}^{2}$ \left\| {{\gamma _2}} \right\|_\infty ^2 \le \left\| {{\cal R}({\gamma _2})} \right\|_\infty ^2 , since γ₂ is PPT and Theorem 1.

Notice that the largest singular value of G_γ₂ is λ₁ by item a) above and the definition of G_γ₂. Hence, by Lemma 3, ‖ℛ(γ₂)‖_∞ = λ₁. Moreover, remind that rank(γ₂) = rank(γ₁) = rank(γ) = k. Therefore, $k λ_{1}^{2} = tr (γ_{2}^{2}) \leq {∥ γ_{2} ∥}_{\infty}^{2} rank (γ_{2}) \leq λ_{1}^{2} . k .$ k\lambda _1^2 = tr(\gamma _2^2) \le \left\| {{\gamma _2}} \right\|_\infty ^2\;{\rm{rank}}({\gamma _2}) \le \lambda _1^2.k.

The inequalities above are, in fact, equalities, which only happens when all the k non-null eigenvalues of γ₂ are equal to λ₁. Therefore, 1 = tr(γ₂) = kλ₁. So $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} . In addition, $1 = tr (γ_{2}) = λ_{1} tr (A_{1}) tr (B_{1}) = \frac{1}{k} \sqrt{k} tr (B_{1}) .$ 1 = tr({\gamma _2}) = {\lambda _1}tr({A_1})tr({B_1}) = {1 \over k}\sqrt k \;tr({B_1}).

So $tr (B_{1}) = \sqrt{k}$ tr({B_1}) = \sqrt k and $tr (B_{1}^{2}) = 1$ tr(B_1^2) = 1 . Recall that $G_{γ_{2}} (\frac{1}{λ_{1}} A_{1}) = B_{1}$ {G_{{\gamma _2}}}\left( {{1 \over {{\lambda _1}}}{A_1}} \right) = {B_1} is a positive semidefinite Hermitian matrix, since G_γ₂is a positive map and $\frac{1}{λ_{1}} A_{1} = \sqrt{k} Id$ {1 \over {{\lambda _1}}}{A_1} = \sqrt k Id . Under these conditions the only possibility for B₁ is $B_{1} = \frac{Id}{\sqrt{k}}$ {B_1} = {{Id} \over {\sqrt k }} . Therefore ${(γ_{2})}_{B} = G_{γ_{2}} (Id) = G_{γ_{2}} (\sqrt{k} A_{1}) = \frac{Id}{k}, {(γ_{2})}_{A} = F_{γ_{2}} (Id) = F_{γ_{2}} (\sqrt{k} B_{1}) = \frac{Id}{k} .$ {({\gamma _2})_B} = {G_{{\gamma _2}}}(Id) = {G_{{\gamma _2}}}(\sqrt k {A_1}) = {{Id} \over k},\;\;\;{({\gamma _2})_A} = {F_{{\gamma _2}}}(Id) = {F_{{\gamma _2}}}(\sqrt k {B_1}) = {{Id} \over k}.

Finally, notice that γ₂ = (R^* ⊗ Id)γ₁(R ⊗ Id) = (R^* ⊗ S)γ(R ⊗ S^*). The proof is complete.

5.

Lower Bound for the Rank of the Special Triad

In this section, we prove that rank(γ) ≥ k, whenever a state γ is PPT or SPC or invariant under realignment and rank(γ_A) = rank(γ_B) = k. Then we show that if rank(γ) = k, then γ is separable in each of these cases.

Explicit examples of states satisfying the hypotheses of these theorems are the separable states $γ_{1} = \sum_{i = 1}^{k} a_{i} a_{i}^{*} \otimes a_{i} a_{i}^{*}, γ_{2} = \sum_{i = 1}^{k} a_{i} a_{i}^{*} \otimes \bar{a_{i}} a_{i}^{t},$ {\gamma _1} = \sum\nolimits_{i = 1}^k {a_i}a_i^* \otimes {a_i}a_i^*,\;\;\;{\gamma _2} = \sum\nolimits_{i = 1}^k {a_i}a_i^* \otimes \overline {{a_i}} a_i^t, where a₁, …, a_k is any basis of ℂ^k. It is clear that their reduced ranks and their ranks are equal to k. In addition, it is clear that γ₁ is positive under partial transpose and $R (γ_{1}^{Γ}) = R (γ_{2}) = γ_{2}$ {\cal R}(\gamma _1^\Gamma ) = {\cal R}({\gamma _2}) = {\gamma _2} . Thus γ₁ is also SPC and γ₂ is invariant under realignment.

We start this section by proving the SPC and the invariant under realignment cases. In their proofs we use the result that guarantees their separability whenever their non-null Schmidt coefficients are equal, which follows from the complete reducibility property [5, Proposition 15].

5.1.

First Cases: SPC States and Invariant under Realignment States

Before proving the inequality, notice that by the symmetry of their Schmidt decompositions [5, Corollary 25], γ_B = γ_A or $γ_{B} = \bar{γ_{A}}$ {\gamma _B} = \overline {{\gamma _A}} , when γ is SPC or invariant under realignment, respectively. Hence rank(γ_A) = rank(γ_B). In addition, we can assume without loss of generality that rank(γ_A) = k, otherwise we would be able to embed γ in ℳ_s ⊗ ℳ_s, where s = rank(γ_A), and obtain the same result.

Theorem 4

If γ ∈ ℳ_k ⊗ ℳ_k is an SPC state such that rank(γ_A) = k, then rank(γ) ≥ k. In addition, if the equality holds, then γ is separable.

Proof

By Theorem 2, there is a invertible matrix R such that $δ = (R^{*} \otimes R^{*}) γ (R \otimes R) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes γ_{i},$ \delta = ({R^*} \otimes {R^*})\gamma (R \otimes R) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes {\gamma _i}, where $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} , $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }} , $tr (γ_{i} γ_{1}) = \frac{tr (γ_{i})}{\sqrt{k}} = 0$ tr({\gamma _i}{\gamma _1}) = {{tr({\gamma _i})} \over {\sqrt k }} = 0 for i > 1. So $δ_{A} = \sum_{ı = 1}^{n} λ_{i} γ_{i} tr (γ_{i}) = \frac{Id}{k}$ {\delta _A} = \sum\limits_{i = 1}^n {{\lambda _i}{\gamma _i}tr({\gamma _i})} = {{Id} \over k} .

Now, notice that δ still is SPC by [5, Corollary 25]. Hence ${∥ δ ∥}_{\infty} \leq {∥ δ_{A} ∥}_{\infty} = \frac{1}{k}$ {\left\| \delta \right\|_\infty } \le {\left\| {\delta _A}\right\|_\infty } = {1 \over k} , by Theorem 1.

So $\frac{1}{k} \geq {∥ δ ∥}_{\infty} \geq \frac{tr (δ)}{rank (δ)} = \frac{1}{rank (δ)}$ {1 \over k} \ge {\left\| \delta \right\|_\infty } \ge {{tr(\delta )} \over {{\rm{rank}}(\delta )}} = {1 \over {{\rm{rank}}(\delta )}} . Hence rank(δ) ≥ k.

Since R is invertible, rank(γ) = rank(δ) ≥ k.

For the next part, assume rank(γ) = k. Then rank(δ) = k. Therefore, $1 = tr (δ) \leq {∥ δ ∥}_{\infty} rank (δ) = \frac{1}{k} . k = 1 .$ 1 = tr(\delta ) \le {\left\| \delta \right\|_\infty }{\rm{rank}}(\delta ) = {1 \over k}.k = 1

Since the equality tr(δ) = ‖δ‖_∞ rank(δ) holds, the non-null eigenvalues of δ are equal to ‖δ‖_∞. So tr(δ) = k‖δ‖_∞ = 1. Hence ${∥ δ ∥}_{\infty} = \frac{1}{k}$ {\left\| \delta \right\|_\infty } = {1 \over k} and $tr (δ^{2}) = \frac{1}{k}$ tr({\delta ^2}) = {1 \over k} .

Next, since the linear contraction – R((⋅)^Γ) – preserves the Frobenius norm of δ, (7) $\frac{1}{k} = tr (δ^{2}) = tr (R (δ^{Γ}) R {(δ^{Γ})}^{*}) \leq {∥ R (δ^{Γ}) ∥}_{\infty} {∥ R (δ^{Γ}) ∥}_{1} .$ {1 \over k} = tr({\delta ^2}) = tr({\cal R}({\delta ^\Gamma }){\cal R}{({\delta ^\Gamma })^*}) \le {\left\| {\cal R}({\delta ^\Gamma })\right\|_\infty }{\left\| {\cal R}({\delta ^\Gamma })\right\|_1}.

By item (5) of Lemma 1, ℛ(δ^Γ) = ℛ(δ)F. Since F is unitary, ‖ℛ(δ)F ‖_∞ = ‖ℛ(δ)‖_∞.

Therefore, ${∥ R (δ^{Γ}) ∥}_{\infty} = {∥ R (δ) F ∥}_{\infty} = {∥ R (δ) ∥}_{\infty} = λ_{1} = \frac{1}{k} .$ {\left\| {\cal R}({\delta ^\Gamma })\right\|_\infty } = {\left\| {\cal R}(\delta )F\right\|_\infty } = {\left\| {\cal R}(\delta )\right\|_\infty } = {\lambda _1} = {1 \over k}.

Now, since δ is SPC, by its definition, ℛ(δ^Γ) is positive semidefinite. Hence ${∥ R (δ^{Γ}) ∥}_{1} \leq {∥ R {(δ^{Γ})}^{Γ} ∥}_{1} .$ {\left\| {\cal R}({\delta ^\Gamma })\right\|_1} \le {\left\| {\cal R}{({\delta ^\Gamma })^\Gamma }\right\|_1}.

By item (7) of Lemma 1, ℛ(δ^Γ)^Γ = δF. Thus, ‖ℛ(δ^Γ)‖₁ ≤ ‖ℛ(δ^Γ)^Γ‖₁ = ‖δF ‖₁ = ‖δ‖₁ = 1.

Using these pieces of information in Equation (7) we obtain (8) $\frac{1}{k} = tr (R (δ^{Γ}) R {(δ^{Γ})}^{*}) \leq {∥ R (δ^{Γ}) ∥}_{\infty} {∥ R (δ^{Γ}) ∥}_{1} \leq \frac{1}{k} . 1 .$ {1 \over k} = tr({\cal R}({\delta ^\Gamma }){\cal R}{({\delta ^\Gamma })^*}) \le {\left\| {\cal R}({\delta ^\Gamma })\right\|_\infty }{\left\| {\cal R}({\delta ^\Gamma })\right\|_1} \le {1 \over k}.1.

Again, tr(ℛ(δ^Γ)²) = ‖ℛ(δ^Γ)‖_∞‖ℛ(δ^Γ)‖₁ only holds if the non-null eigenvalues of the positive semidefinite Hermitian matrix ℛ(δ^Γ) are equal to ${∥ R (δ^{Γ}) ∥}_{\infty} = λ_{1} = \frac{1}{k}$ {\left\| {\cal R}({\delta ^\Gamma })\right\|_\infty } = {\lambda _1} = {1 \over k} .

Finally, the non-null eigenvalues of ℛ(δ^Γ) are the singular values of G_γ, which are the non-null Schmidt coefficients of δ, as explained in the proof of Lemma 3. Since δ is SPC and its non-null Schmidt coefficients are equal, δ is separable by [5, Proposition 15]. Since R is invertible, γ is separable too.

The invariant under realignment counterpart is proved next in a similar way with minor modifications.

Theorem 5

If γ ∈ ℳ_k ⊗ ℳ_k is an invariant under realignment state such that rank(γ_A) = k, then rank(γ) ≥ k. In addition, if the equality holds, then γ is separable.

Proof

By Theorem 2, there is a invertible matrix R such that $δ = (R^{*} \otimes R^{t}) γ (R \otimes \bar{R}) = \sum_{i = 1}^{n} λ_{i} γ_{i} \otimes \bar{γ_{i}},$ \delta = ({R^*} \otimes {R^t})\gamma (R \otimes \overline R ) = \sum\nolimits_{i = 1}^n {\lambda _i}{\gamma _i} \otimes \overline {{\gamma _i}} , where $λ_{1} = \frac{1}{k}$ {\lambda _1} = {1 \over k} , $γ_{1} = \frac{Id}{\sqrt{k}}$ {\gamma _1} = {{Id} \over {\sqrt k }} , $tr (γ_{i} γ_{1}) = \frac{tr (γ_{i})}{\sqrt{k}} = 0$ tr({\gamma _i}{\gamma _1}) = {{tr({\gamma _i})} \over {\sqrt k }} = 0 for i > 1. So $δ_{A} = \sum_{ı = 1}^{n} λ_{i} γ_{i} tr (γ_{i}) = \frac{Id}{k}$ {\delta _A} = \sum\limits_{i = 1}^n {{\lambda _i}{\gamma _i}tr({\gamma _i})} = {{Id} \over k} .

Now, by item (3) of Lemma 1, $R (δ) = R ((R^{*} \otimes R^{t}) γ (R \otimes \bar{R})) = (R^{*} \otimes R^{t}) R (γ) (R \otimes \bar{R}) = (R^{*} \otimes R^{t}) γ (R \otimes \bar{R}) = δ .$ {\cal R}(\delta ) = {\cal R}(({R^*} \otimes {R^t})\gamma (R \otimes \overline R )) = ({R^*} \otimes {R^t}){\cal R}(\gamma )(R \otimes \overline R ) = ({R^*} \otimes {R^t})\gamma (R \otimes \overline R ) = \delta .

Thus, δ is invariant under realignment. Hence ${∥ δ ∥}_{\infty} \leq {∥ δ_{A} ∥}_{\infty} = \frac{1}{k}$ {\left\| \delta \right\|_\infty } \le {\left\| {\delta _A}\right\|_\infty } = {1 \over k} , by Theorem 1.

So $\frac{1}{k} \geq {∥ δ ∥}_{\infty} \geq \frac{tr (δ)}{rank (δ)} = \frac{1}{rank (δ)}$ {1 \over k} \ge {\left\| \delta \right\|_\infty } \ge {{tr(\delta )} \over {{\rm{rank}}(\delta )}} = {1 \over {{\rm{rank}}(\delta )}} . Hence rank(δ) ≥ k.

Since R is invertible, rank(γ) = rank(δ) ≥ k.

For the next part, assume rank(γ) = k. Then rank(δ) = k. Therefore, $1 = tr (δ) \leq {∥ δ ∥}_{\infty} rank (δ) = \frac{1}{k} . k = 1$ 1 = tr(\delta ) \le {\left\| \delta \right\|_\infty }\;{\rm{rank}}(\delta ) = {1 \over k}.k = 1.

Since the equality tr(δ) = ‖δ‖_∞ rank(δ) holds, the non-null eigenvalues of δ are equal to ‖δ‖_∞. Moreover, tr(δ) = k‖δ‖_∞ = 1. Hence ${∥ δ ∥}_{\infty} = \frac{1}{k}$ {\left\| \delta \right\|_\infty } = {1 \over k} .

Since ℛ(δ) = δ, the non-null singular values of ℛ(δ) are the non-null eigenvalues of δ, which are equal in this case. Hence the non-null Schmidt coefficients of the Schmidt decomposition of δ are equal by Lemma 3. We know that every invariant under realignment state with equal non-null Schmidt coefficients is separable by [5, Proposition 15]. So δ is separable and so is γ, since R is invertible.

5.2.

The PPT Counterpart

For our final results, we need some tools developed in Sections 2 and 3 together with the complete reducibility property. First, we show that the rank of a PPT state is greater or equal to its reduced ranks in the next lemma.

Lemma 8

Let γ ∈ ℳ_k ⊗ ℳ_m be a PPT state. Then rank(γ) ≥ max{rank(γ_A), rank(γ_B)}.

Proof

Let us assume without loss of generality that max{rank(γ_A), rank(γ_B)} = rank(γ_A) = k. So there is an invertible matrix R ∈ ℳ_k such that $R γ_{A} R^{*} = \frac{1}{k} Id$ R{\gamma _A}{R^*} = {1 \over k}Id . Define δ = (R ⊗ Id)γ(R^* ⊗ Id) and notice that $δ_{A} = \frac{1}{k} Id$ {\delta _A} = {1 \over k}Id .

Since δ is PPT, by Theorem 1, ${∥ δ ∥}_{\infty} \leq {∥ δ_{A} ∥}_{\infty} = \frac{1}{k}$ {\left\| \delta \right\|_\infty } \le {\left\| {\delta _A}\right\|_\infty } = {1 \over k} . Hence $1 = tr (δ_{A}) = tr (δ) \leq {∥ δ ∥}_{\infty} rank (δ) \leq \frac{1}{k} rank (δ) .$ 1 = tr({\delta _A}) = tr(\delta ) \le {\left\| \delta \right\|_\infty }\;{\rm{rank}}(\delta ) \le {1 \over k}{\rm{rank}}(\delta ).

Thus, rank(γ) = rank(δ) ≥ max{rank(γ_A), rank(γ_B)}.

We complete this paper by proving that a PPT state with minimal rank must be separable which is the PPT counterpart of the Theorems 4 and 5.

Theorem 6

If δ ∈ ℳ_k ⊗ ℳ_k is a PPT state such that rank(δ) = rank(δ_A) = rank(δ_B) = k, then δ is separable.

Proof

This result is trivial in ℳ₂ ⊗ ℳ₂, since every PPT state there is separable. Assume the result is true in ℳ_i ⊗ ℳ_i for i < k. Let us prove the result in ℳ_k ⊗ ℳ_k.

First of all, by Theorem 3, there are invertible matrices M, N ∈ ℳ_k such that γ = (M ⊗N )δ(M^* ⊗N^*) satisfies $γ_{A} = γ_{B} = \frac{1}{k} Id$ {\gamma _A} = {\gamma _B} = {1 \over k}Id . Notice that γ is also PPT and its rank is equal to k.

Next, by Theorem 1, since ${∥ γ ∥}_{\infty} \leq ∥ γ_{A} ∥ = \frac{1}{k}$ {\left\| \gamma \right\|_\infty } \le \left\| {{\gamma _A}} \right\| = {1 \over k} and $1 = tr (γ_{A}) = tr (γ) \leq {∥ γ ∥}_{\infty} rank (γ) \leq \frac{1}{k} k = 1,$ 1 = tr({\gamma _A}) = tr(\gamma ) \le {\left\| \gamma \right\|_\infty }\;{\rm{rank}}(\gamma ) \le {1 \over k}k = 1, we obtain all k non-null eigenvalues of γ equal to $\frac{1}{k}$ {1 \over k} .

Now, since γ is a positive semidefinite Hermitian matrix, the linear transformations F_γ and G_γ are positive maps and adjoints with respect to the trace inner product.

In addition, notice that $\frac{Id}{k} = γ_{A} = F_{γ} (Id)$ {{Id} \over k} = {\gamma _A} = {F_\gamma }(Id) and $\frac{Id}{k} = γ_{B} = G_{γ} (Id)$ {{Id} \over k} = {\gamma _B} = {G_\gamma }(Id) . Hence $F_{γ} (G_{γ} (Id)) = \frac{1}{k^{2}} Id$ {F_\gamma }({G_\gamma }(Id)) = {1 \over {{k^2}}}Id .

By [24, Theorem 2.3.7], the spectral radius of the positive operator F_γ ∘ G_γ is $\frac{1}{k^{2}}$ {1 \over {{k^2}}} . Hence the largest singular value of G_γ and F_γ is $\frac{1}{k}$ {1 \over k} , since they are adjoints. Thus, ${∥ G_{γ} (X) ∥}_{2} \leq \frac{1}{k} and {∥ F_{γ} (X) ∥}_{2} \leq \frac{1}{k}, whenever {∥ X ∥}_{2} = 1 .$ {\left\| {G_\gamma }(X)\right\|_2} \le {1 \over k}\;{\rm{and}}\;{\left\| {F_\gamma }(X)\right\|_2} \le {1 \over k},\;{\rm{whenever}}\,{\left\| X\right\|_2} = 1.

Next, since γ has k linearly independent eigenvectors associated to $\frac{1}{k}$ {1 \over k} , by combining 2 of them we can find an eigenvector v ∈ ℂ^k ⊗ ℂ^k associated to $\frac{1}{k}$ {1 \over k} such that ‖v‖₂ = 1 and rank(v) = m < k.

Notice that there are R, S ∈ ℳ_k with rank m such that $v = (R \otimes S) u (u as defined in Remark 1) and {{RR}^{*}‖}_{2} = {{SS}^{*}‖}_{2} = 1.$ v = \left( {R \otimes S} \right)u\left( {u\;{\rm{as}}\;{\rm{defined}}\;{\rm{in}}\;{\rm{Remark}}\;1} \right)\;{\rm{and}}\;{\left\| {R{R^*}} \right\|_2} = {\left\| {S{S^*}} \right\|_2} = 1. $\begin{array}{l} Therefore \frac{1}{k} & = tr (γ v v^{*}) \\ = tr ((R^{*} \otimes S^{*}) γ (R \otimes S) u u^{t}) \\ = tr ((R^{*} \otimes S^{t}) γ^{Γ} (R \otimes \bar{S}) F) (F as defined in item c) of Remark 1) \\ \leq tr ((R^{*} \otimes S^{t}) γ^{Γ} (R \otimes \bar{S})) (since γ^{Γ} and Id - F are positive semidefinite) \\ = tr (γ (R R^{*} \otimes S S^{*})) \\ = tr (G_{γ} (R R^{*}) S S^{*}) \\ \leq {∥ G_{γ} (R R^{*}) ∥}_{2} {∥ S S^{*} ∥}_{2} \leq \frac{1}{k} . 1 (since the largest singular value of G_{γ} is \frac{1}{k}) \end{array}$ \matrix{ {{\rm{Therefore}}\;{1 \over k}} \hfill & { = tr(\gamma v{v^*})} \hfill \cr {} \hfill & { = tr(({R^*} \otimes {S^*})\gamma (R \otimes S)u{u^t})} \hfill \cr {} \hfill & { = tr(({R^*} \otimes {S^t}){\gamma ^\Gamma }(R \otimes \overline S )F)\;(F\;{\rm{as}}\;{\rm{defined}}\;{\rm{in}}\;{\rm{item}}\;c)\;{\rm{of}}\;{\rm{Remark}}\;1)} \hfill \cr {} \hfill & { \le tr(({R^*} \otimes {S^t}){\gamma ^\Gamma }(R \otimes \overline S ))\;({\rm{since}}\;{\gamma ^\Gamma }\;{\rm{and}}\;Id - F\;{\rm{are}}\;{\rm{positive}}\;{\rm{semidefinite}})} \hfill \cr {} \hfill & { = tr(\gamma (R{R^*} \otimes S{S^*}))} \hfill \cr {} \hfill & { = tr({G_\gamma }(R{R^*})S{S^*})} \hfill \cr {} \hfill & { \le {\left\| {G_\gamma }(R{R^*})\right\|_2}{\left\| S{S^*}\right\|_2} \le {1 \over k}.1\;\left( {{\rm{since}}\;{\rm{the}}\;{\rm{largest}}\;{\rm{singular}}\;{\rm{value}}\;{\rm{of}}\;{G_\gamma }\;{\rm{is}}\;{1 \over k}} \right)} \hfill \cr }

Therefore, all the inequalities above are equalities, which imply

(1)
G_γ (RR^*) = λSS^* for some λ > 0, since G_γ is a positive map, and
(2)
$\frac{1}{k} = {∥ G_{γ} (R R^{*}) ∥}_{2} = λ {∥ S S^{*} ∥}_{2} = λ$ {1 \over k} = \;{\left\| {G_\gamma }(R{R^*})\right\|_2} = \lambda {\left\| S{S^*}\right\|_2} = \lambda .

Hence $G_{γ} (R R^{*}) = \frac{1}{k} S S^{*}$ {G_\gamma }(R{R^*}) = {1 \over k}S{S^*} . Analogously, we get $F_{γ} (S S^{*}) = \frac{1}{k} R R^{*}$ {F_\gamma }(S{S^*}) = {1 \over k}R{R^*} , since tr(γ(RR^*⊗SS^*)) = tr(RR^*F_γ (SS^*)).

Therefore, $F_{γ} (G_{γ} (R R^{*})) = \frac{1}{k^{2}} R R^{*}$ {F_\gamma }({G_\gamma }(R{R^*})) = {1 \over {{k^2}}}R{R^*} and rank(RR^*) = m < k.

Since γ is PPT, by the complete reducibility property, (9) $γ = (V \otimes W) γ (V \otimes W) + (V^{⊥} \otimes W^{⊥}) γ (V^{⊥} \otimes W^{⊥}),$ \gamma = (V \otimes W)\gamma (V \otimes W) + ({V^ \bot } \otimes {W^ \bot })\gamma ({V^ \bot } \otimes {W^ \bot }), where V, W, V^⊥, W^⊥ are orthogonal projections onto Im(RR^*), Im(SS^*), ker(RR^*) and ker(SS^*), respectively.

The proof is almost done, we just need to check that (V ⊗ W )γ(V ⊗ W ), (V^⊥ ⊗ W^⊥)γ(V^⊥ ⊗ W^⊥) are multiples of states satisfying the hypotheses and since rank(V ) = rank(W ) < k and rank(V^⊥) = rank(W^⊥) < k, the result follows by induction.

Notice that by Equation (9) and the definition of G_γ, $\begin{array}{l} Im (G_{γ} (V)) \subset Im (W), Im (G_{γ} (V^{⊥})) \subset Im (W^{⊥}) and \\ Im (F_{γ} (W)) \subset Im (V), Im (F_{γ} (W^{⊥})) \subset Im (V^{⊥}) . \end{array}$ \matrix{ {{\rm{Im}}({G_\gamma }(V)) \subset {\rm{Im}}(W),{\rm{Im}}({G_\gamma }({V^ \bot })) \subset {\rm{Im}}({W^ \bot })\;{\rm{and}}} \hfill \cr {{\rm{Im}}({F_\gamma }(W)) \subset {\rm{Im}}(V),{\rm{Im}}({F_\gamma }({W^ \bot })) \subset {\rm{Im}}({V^ \bot }).\quad } \hfill \cr }

Next, recall that V + V^⊥ = W + W^⊥ = Id, V V^⊥ = W W^⊥ = 0 and $\begin{array}{l} G_{γ} (V) + G_{γ} (V^{⊥}) = G_{γ} (Id) = \frac{1}{k} Id = \frac{1}{k} W + \frac{1}{k} W^{⊥}, \\ F_{γ} (W) + F_{γ} (W^{⊥}) = F_{γ} (Id) = \frac{1}{k} Id = \frac{1}{k} V + \frac{1}{k} V^{⊥} . \end{array}$ \matrix{ {{G_\gamma }(V) + {G_\gamma }({V^ \bot }) = {G_\gamma }(Id) = {1 \over k}Id = {1 \over k}W + {1 \over k}{W^ \bot },} \hfill \cr {{F_\gamma }(W) + {F_\gamma }({W^ \bot }) = {F_\gamma }(Id) = {1 \over k}Id = {1 \over k}V + {1 \over k}{V^ \bot }.} \hfill \cr }

Therefore $G_{γ} (V) = \frac{1}{k} W, F_{γ} (W) = \frac{1}{k} V and G_{γ} (V^{⊥}) = \frac{1}{k} W^{⊥}, F_{γ} (W^{⊥}) = \frac{1}{k} V^{⊥} .$ {G_\gamma }(V) = {1 \over k}W,\;{F_\gamma }(W) = {1 \over k}V\;{\rm{and}}\;{G_\gamma }({V^ \bot }) = {1 \over k}{W^ \bot },{F_\gamma }({W^ \bot }) = {1 \over k}{V^ \bot }.

Now, define $γ_{1} = \frac{k}{m} (V \otimes W) γ (V \otimes W) and γ_{2} = \frac{k}{k - m} (V^{⊥} \otimes W^{⊥}) γ (V^{⊥} \otimes W^{⊥}) .$ {\gamma _1} = {k \over m}(V \otimes W)\gamma (V \otimes W)\;{\rm{and}}\;{\gamma _2} = {k \over {k - m}}({V^ \bot } \otimes {W^ \bot })\gamma ({V^ \bot } \otimes {W^ \bot }).

Notice that ${(γ_{1})}_{A} = F_{γ_{1}} (Id) = \frac{k}{m} F_{γ} (W) = \frac{1}{m} V and {(γ_{1})}_{B} = G_{γ_{1}} (Id) = \frac{k}{m} G_{γ} (V) = \frac{1}{m} W .$ {({\gamma _1})_A} = {F_{{\gamma _1}}}(Id) = {k \over m}{F_\gamma }(W) = {1 \over m}V\;{\rm{and}}\;{({\gamma _1})_B} = {G_{{\gamma _1}}}(Id) = {k \over m}{G_\gamma }(V) = {1 \over m}W.

Thus, max{rank((γ₁)_A), rank((γ₁)_B)} = max{rank(V ), rank(W )} = max{rank(R), rank(S)} = m.

Moreover, notice that ${(γ_{2})}_{A} = F_{γ_{2}} (Id) = \frac{k}{k - m} F_{γ} (W^{⊥}) = \frac{1}{k - m} V^{⊥} and {(γ_{2})}_{B} = G_{γ_{2}} (Id) = \frac{k}{k - m} G_{γ} (V^{⊥}) = \frac{1}{k - m} W^{⊥} .$ {({\gamma _2})_A} = {F_{{\gamma _2}}}(Id) = {k \over {k - m}}{F_\gamma }({W^ \bot }) = {1 \over {k - m}}{V^ \bot }\;{\rm{and}}\;{({\gamma _2})_B} = {G_{{\gamma _2}}}(Id) = {k \over {k - m}}{G_\gamma }({V^ \bot }) = {1 \over {k - m}}{W^ \bot }.

Therefore max{rank((γ₂)_A), rank((γ₂)_B)} = max{rank(V^⊥), rank(W^⊥)} = k − m.

By their definitions, γ₁ and γ₂ are PPT. So, by Lemma 8, rank(γ₁) ≥ m and rank(γ₂) ≥ k − m.

Recall that k = rank(γ) = rank(γ₁) + rank(γ₂) ≥ m + (k − m). Thus rank(γ₁) = m and rank(γ₂) = k − m.

Since $γ = \frac{m}{k} γ_{1} + \frac{k - m}{k} γ_{2}$ \gamma = {m \over k}{\gamma _1} + {{k - m} \over k}{\gamma _2} , γ has k eigenvalues equal to $\frac{1}{k}$ {1 \over k} and γ₁γ₂ = 0,

γ₁ has m eigenvalues equal to $\frac{1}{m}$ {1 \over m} and the others 0,
γ₂ has k − m eigenvalues equal to $\frac{1}{k - m}$ {1 \over {k - m}} and the others 0.

Hence,

γ₁ has m eigenvalues equal to $\frac{1}{m}$ {1 \over m} , ${(γ_{1})}_{A} = \frac{1}{m} V$ {({\gamma _1})_A} = {1 \over m}V , ${(γ_{1})}_{B} = \frac{1}{m} W$ {({\gamma _1})_B} = {1 \over m}W and rank(V ) = rank(W ) = m.
γ₂ has k − m eigenvalues equal to $\frac{1}{k - m}$ {1 \over {k - m}} , ${(γ_{2})}_{A} = \frac{1}{k - m} V^{⊥}$ {({\gamma _2})_A} = {1 \over {k - m}}{V^ \bot } , ${(γ_{1})}_{B} = \frac{1}{k - m} W^{⊥}$ {({\gamma _1})_B} = {1 \over {k - m}}{W^ \bot } and rank(V^⊥) = rank(W^⊥) = k − m.

By induction hypothesis, γ₁ and γ₂ are separable and so is γ and δ.

6.

Summary and Conclusions

In this paper we proved new results for a triad of quantum state types which includes the positive under partial transpose type. We obtained the same upper bound for the spectral radius of these types of quantum states. We then showed that the first two types can be put in the filter normal form and the third type only under some restriction. Finally, we proved that there is a lower bound for their ranks and whenever this lower bound is attained these states are separable. This last result is another consequence of their complete reducibility property. This is sufficient evidence that these states are deeply connected. Moreover, their complete reducibility property is a unifying force connecting and providing many results in entanglement theory.

A Triality Pattern in Entanglement Theory

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