Nullspaces of entanglement breaking channels and applications

Quantum entanglement theory lies at the heart of many topics in modern quantum information science. On the one hand, quantum entanglement acts as a crucial resource in many applications of quantum information, such as quantum algorithms and quantum cryptography, whereas on the other hand, it is still a somewhat mysterious and not-so-well understood phenomena. One of the deepest and most challenging topics in entanglement theory is the analysis of how quantum operations behave when acting on entangled states; in particular, on the possible convertibility or destruction of entanglement.

For almost two decades, a special class of quantum operations, those that 'break' entanglement through their actions on quantum states, have been heavily investigated. Original motivations for the consideration of these so-called entanglement breaking channels were mainly information theoretic, with emphasis on quantum channel capacity investigations, where the class has played a key role in such investigations and yielded surprizing information theoretic results for quantum channels. It turns out that many important and well-studied examples of quantum channels are entanglement breaking, and this class of maps has now arisen in numerous areas of the subject. We point the reader to the seminal papers on the topic [18, 21] and forward references for an entrance into the extensive literature on these channels.

In this paper, we contribute to the study of entanglement breaking channels first through an investigation into their nullspace structures, and then by deriving a pair of related applications to different areas of quantum information. Specifically, we first show that every self-adjoint operator space of trace zero matrices is the nullspace of such a channel. Building on this, and taking motivation from quantum privacy, we derive a test for mixed unitarity of quantum channels [3, 16, 17, 22, 32] based on entanglement breaking channel nullspaces and complementary channel [19–21] behaviour. Starting from a connection with channel nullspaces, we also identify conditions that guarantee the existence of private algebras [2, 5, 6, 9, 13, 14, 23, 26, 33] for certain classes of entanglement breaking channels based on an analysis of multiplicative domains from operator theory [10, 11, 24, 29, 39, 40] for the channels.

This paper is organized as follows. The next section includes preliminary material. In section 3 we give the operator space nullspace construction. Section 4 includes the derivation of the mixed unitary test. Then we present the identification and construction of private algebras in section 5.

Quantum channels are central objects of study in quantum information [19, 35] and are given mathematically by completely positive and trace preserving maps on (in the finite-dimensional case) the set of complex n × n matrices ${M}_{n}\left(\mathbb{C}\right)$. Every channel ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ can be represented in the operator-sum form by a set of operators ${V}_{i}\in {M}_{n}\left(\mathbb{C}\right)$, such that ${\Phi}\left(\rho \right)={\sum }_{i}{V}_{i}\rho {V}_{i}^{{\ast}}$ and with the trace-preservation condition ${\sum }_{i}{V}_{i}^{{\ast}}{V}_{i}=I$ satisfied where I is the identity matrix. The dual map Φ^† on ${M}_{n}\left(\mathbb{C}\right)$ will also arise in our analysis, which is the completely positive (and unital when Φ is trace preserving) map given by ${{\Phi}}^{{\dagger}}\left(X\right)={\sum }_{i}{V}_{i}^{{\ast}}X{V}_{i}$.

When convenient we will view elements of ${M}_{n}\left(\mathbb{C}\right)$ as matrix representations of the set of operators acting on n-dimensional Hilbert space $\mathcal{H}={\mathbb{C}}^{n}$, represented in the standard orthonormal basis {e₁, ..., e_n }. Outer products will be written as rank one operators vw* for $v,w\in \mathcal{H}$, defined by (v w*)(u) = (w*u)v, where w*u is the inner product of w with u. Note the implication that for us, inner products are antilinear in their first argument, not the second. Additionally we will use the default notation ρ for density operators or matrices; that is, positive operators with trace equal to one. We will also use the notation ${M}_{n}{\left(\mathbb{C}\right)}_{0}$ to denote the set of trace zero n × n complex matrices.

2.1. Entanglement breaking channels

An important class of channels are those that break all entanglement when acting on a composite system with the identity channel of the same size, Φ ⊗ id [18, 21]. There are numerous equivalent characterizations of entanglement breaking channels, including a physically motivated description as the composition of quantum–classical and classical-quantum channels in the same orthonormal basis. The Holevo form for such channels [18] is given as follows.

Definition 2.1. A quantum channel ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ is entanglement breaking if it can be written as:

$$\begin{equation}{\Phi}\left(\rho \right)=\sum\limits _{k=1}^{d}\mathrm{Tr}\left({F}_{k}\rho \right){R}_{k},\end{equation} \tag{ 1 }$$

with the {F_k } forming a positive-operator valued measure (F_k ⩾ 0 and ∑_k F_k = I) and each R_k a density operator. We shall make the further assumption that none of the F_k are zero, which has always been followed in practice.

We will also make use of the characterization of entanglement breaking channels as precisely the channels with an operator-sum representation comprised of rank one Kraus operators. That is, Φ is entanglement breaking if and only if there are rank one operators ${\left\{{v}_{i}{w}_{i}^{{\ast}}\right\}}_{i=1}^{d}$ such that

$$\begin{equation}{\Phi}\left(\rho \right)=\sum\limits _{i=1}^{d}{v}_{i}{w}_{i}^{{\ast}}\enspace \rho \enspace {w}_{i}{v}_{i}^{{\ast}}.\end{equation} \tag{ 2 }$$

Without loss of generality, we will assume throughout that ||v_i || = 1 for all i, and hence trace preservation gives the constraint: ${\sum }_{i}{w}_{i}{w}_{i}^{{\ast}}=I$. To avoid degeneracy we also assume each w_i ≠ 0.

In section 4 we will consider a slightly more general class of entanglement breaking maps, still using the rank one form of equation (2), in the development of the mixed unitary test presented there.

2.2. Complementary channels

The following notion will be used in two of our sections below. The concept of complementary quantum channels was first explored in detail in quantum information as part of investigations into fundamental conjectures on multiplicativity and additivity of entropy and related channel capacity problems. The study of such complementarity expanded into other areas, including analysis of the information-disturbance trade-off for channels and the complementarity of quantum error-correcting codes with private quantum codes. For more background on complementary channels see [19–21] and forward references.

Definition 2.2. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be a quantum channel with a minimal set of Kraus operators ${\left\{{V}_{i}\right\}}_{i=1}^{d}$. The canonical complement of Φ is the channel ${{\Phi}}^{\mathrm{C}}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{d}\left(\mathbb{C}\right)$ defined by

$$\begin{equation}{{\Phi}}^{\mathrm{C}}\left(\rho \right)=\sum\limits _{i,j=1}^{d}\mathrm{Tr}\left({V}_{j}^{{\ast}}{V}_{i}\rho \right){E}_{ij},\end{equation} \tag{ 3 }$$

with ${E}_{ij}={e}_{i}{e}_{j}^{{\ast}}$ for 1 ⩽ i, j ⩽ n.

A complementary channel for Φ is any isometric adjunction of the canonical complement; that is, Ψ is a complementary channel for Φ if and only if there exists an isometry W such that

$$\begin{equation*}{\Psi}\left(\rho \right)=W{{\Phi}}^{\mathrm{C}}\left(\rho \right){W}^{{\ast}}.\end{equation*}$$

Complementary channels arise from the Stinespring representation [42] of a channel. The freedom to conjugate by an isometry comes from the inherent freedom to choose a Stinespring representation; alternatively, as any set of Kraus operators ${\left\{\tilde {{V}_{i}}\right\}}_{i=1}^{r}$ for Φ is related to the canonical minimal choice by $\tilde {{V}_{i}}={\sum }_{j=1}^{d}{w}_{ij}{V}_{j}$ for some isometry W = (w_ij ), we see that adjunction by an isometry corresponds simply to picking a different set of Kraus operators for Φ.

Though not directly relevant to our analysis, it is worth noting that the complementary channel of an entanglement breaking channel is a Schur product channel [19, 36], which have also been recently explored [34] in the quantum privacy context to which we now turn.

2.3. Private subspaces and algebras, and channel nullspaces

In this section, we give a construction of entanglement breaking channels that annihilate prescribed operator spaces and discuss a pair of examples.

Note first that the nullspace $\left\{X\in {M}_{n}\left(\mathbb{C}\right):{\Phi}\left(X\right)=0\right\}$ of any quantum channel ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$, in particular as a trace-preserving map, is contained inside the operator subspace of trace zero matrices, and that it is a self-adjoint subspace as a channel is a positive map.

Proposition 3.1. Let $\mathcal{N}$ be a self-adjoint subspace of the trace zero matrices inside ${M}_{n}\left(\mathbb{C}\right)$. Then there is an entanglement breaking channel Φ: ${M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ such that $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\left({\Phi}\right)=\mathcal{N}$.

Proof. In the case that $\mathcal{N}={M}_{n}{\left(\mathbb{C}\right)}_{0}$, we may use the so-called completely depolarizing channel ${{\Phi}}_{\text{CD}}\left(A\right)=\frac{\mathrm{Tr}\left(A\right)}{n}I$, which is evidently in the Holevo form with F₁ = I, ${R}_{1}=\frac{1}{n}I$. (It is also implemented with Kraus operators given by any complete set of rank one matrix units ${E}_{ij}={e}_{i}{e}_{j}^{{\ast}}$, where {e_i } form an orthonormal basis for ${\mathbb{C}}^{n}$.)

Thus, for the rest of the proof assume $\mathcal{N}{\subsetneq}{M}_{n}{\left(\mathbb{C}\right)}_{0}$, and let ${\left\{{H}_{j}\right\}}_{j=1}^{m}$ be a basis of Hermitian operators for ${\mathcal{N}}^{\perp }\cap {M}_{n}{\left(\mathbb{C}\right)}_{0}$ scaled so that ${\Vert}{H}_{j}{\Vert}=\frac{1}{3m}$ for all j. Then both ${F}_{j}={H}_{j}+\frac{1}{2m}I$ and ${F}_{j+m}=-{H}_{j}+\frac{1}{2m}I$ are positive definite for all j and hence ${\left\{{F}_{k}\right\}}_{k=1}^{2m}$ is a POVM.

Now let ${\left\{{R}_{k}\right\}}_{k=1}^{2m}$ be a set of linearly independent density operators inside ${M}_{n}\left(\mathbb{C}\right)$, and define a channel ${\Phi}\left(\rho \right)={\sum }_{k=1}^{2m}\mathrm{Tr}\left(\rho {F}_{k}\right){R}_{k}$, which is evidently entanglement breaking. Then we have X ∈ nullspace(Φ) if and only if Tr(XF_k ) = ⟨X, F_k ⟩ = 0 for all 1 ⩽ k ⩽ 2m. However, we also have by construction:

$$\begin{align*}\hfill {\left(\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{F}_{k}\right\}}_{k=1}^{2m}\right)}^{\perp }& ={\left(\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{H}_{k}\right\}}_{k=1}^{m}\cup \left\{I\right\}\right)}^{\perp }\hfill \\ \hfill & ={\left({\mathcal{N}}^{\perp }\cap {M}_{n}{\left(\mathbb{C}\right)}_{0}\right)}^{\perp }\cap {M}_{n}{\left(\mathbb{C}\right)}_{0}\hfill \\ \hfill & =\mathcal{N},\hfill \end{align*}$$

and so the result follows as $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\left({\Phi}\right)=\mathcal{N}$. □

The following is a simple illustrative example of the construction above.

Example 3.2. The completely depolarizing channel ${{\Phi}}_{\text{CD}}\left(A\right)=\frac{\mathrm{Tr}\left(A\right)}{2}I$ on ${M}_{2}\left(\mathbb{C}\right)$ is also implemented as a mixed unitary channel (discussed in more detail in the next section) with Kraus operators given by the normalized identity and Pauli operators $\left\{\frac{1}{2}I,\frac{1}{2}X,\frac{1}{2}Y,\frac{1}{2}Z\right\}$, where X = E₁₂ + E₂₁, Y = iE₂₁ − iE₁₂, and Z = E₁₁ − E₂₂. Note these three operators also form an orthogonal basis for ${M}_{2}{\left(\mathbb{C}\right)}_{0}$.

If we consider the subspace $\mathcal{N}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{Z\right\}$, then following the construction we can choose ${H}_{1}=\frac{1}{6}X$ and ${H}_{2}=\frac{1}{6}Y$ as Hermitian operators forming a basis for ${\mathcal{N}}^{\perp }\cap {M}_{2}{\left(\mathbb{C}\right)}_{0}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{X,Y\right\}$, This gives us ${F}_{1}=\frac{1}{4}I+\frac{1}{6}X$, ${F}_{2}=\frac{1}{4}I+\frac{1}{6}Y$, ${F}_{3}=\frac{1}{4}I-\frac{1}{6}X$, and ${F}_{4}=\frac{1}{4}I-\frac{1}{6}Y$. Finally, we can take {R₁, R₂, R₃, R₄} to be any set of three linearly independent density operators inside ${M}_{2}\left(\mathbb{C}\right)$, and define Φ as the entanglement breaking channel with {F_k , R_k } defining its Holevo form. One can verify directly that $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\left({\Phi}\right)=\mathcal{N}$.

If we further consider the subspace $\mathcal{N}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{X,Z\right\}$, then in this case the construction gives us ${H}_{1}=\frac{1}{3}Y$ as a Hermitian operator forming an basis for ${\mathcal{N}}^{\perp }\cap {M}_{2}{\left(\mathbb{C}\right)}_{0}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{Y\right\}$, and hence ${F}_{1}=\frac{1}{2}I+\frac{1}{3}Y$ and ${F}_{2}=\frac{1}{2}I-\frac{1}{3}Y$. As above and in the proof, a channel Φ that satisfies $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\left({\Phi}\right)=\mathcal{N}$ can then be explicitly defined in the Holevo form by choosing any two linearly independent density operators {R₁, R₂}.

3.1. Bi-unitary channels

In the spirit of these channel nullspace investigations, though somewhat outside our entanglement breaking channel focus, we can also consider the class of bi-unitary channels. Such channels are described by scenarios in which a system is exposed to unitary noise with some fixed probability 0 < p < 1; as a channel this is given by the map Φ_U (ρ) = (1 − p)ρ + p UρU* for some fixed unitary operator U. These are a special case of mixed unitary channels considered in the next section and have been investigated in quantum error correction and numerical range theory [12].

Suppose A is a non-zero Hermitian matrix in the nullspace of Φ_U . Then we will have

$$\begin{equation*}UA{U}^{{\ast}}=-\frac{1-p}{p}A.\end{equation*}$$

As UAU* has the same spectrum as A, it follows that this equation cannot be satisfied for $p\ne \frac{1}{2}$ and hence nullspace(Φ_U ) = {0} in those cases.

When $p=\frac{1}{2}$, we have a further equation

$$\begin{equation*}UA{U}^{{\ast}}=-A,\end{equation*}$$

which forces A and −A to have the same eigenvalues. Next, we can diagonalize U as

$$\begin{equation*}U=\sum\limits _{i}{w}_{i}{u}_{i}{u}_{i}^{{\ast}},\end{equation*}$$

with the w_i lying on the unit circle of the complex plane and u_i a set of orthonormal eigenvectors for U. Expanding A = (a_ij ) in this basis gives

$$\begin{equation*}UA{U}^{{\ast}}=\sum\limits _{i,j}{w}_{i}{\bar{w}}_{j}{a}_{ij}{u}_{i}{u}_{j}^{{\ast}},\end{equation*}$$

and so ${w}_{i}{\bar{w}}_{j}{a}_{ij}=-{a}_{ij}$ for all i, j.

We thus end up with two options for each entry: a_ij = 0 or w_i = −w_j . This tells us that, in the case $p=\frac{1}{2}$, we have a non-trivial null space for Φ_U determined by the eigenvalues of U that come in phase flip pairs, reminiscent of quantum properties that generate, for instance, the Pauli matrices.

One useful application of the ideas above is to the type of channel known as mixed unitary (or random unitary) channels.

Definition 4.1. A channel ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ is said to be mixed unitary if it can be written in the form

$$\begin{equation*}{\Phi}\left(X\right)=\sum\limits _{i=1}^{d}{p}_{i}{U}_{i}X{U}_{i}^{{\ast}},\end{equation*}$$

where p_i form a probability distribution (p_i > 0, ∑_i p_i = 1) and U_i ∈ U(n) are unitaries. The Kraus operators for Φ are thus given by $\sqrt{{p}_{i}}{U}_{i}$.

The class of mixed unitary channels arise in all areas of quantum information, and so a number of investigations have been conducted on determining when a channel has this form. Important recent works on the topic include a proof that detecting mixed unitarity is NP-hard in general [32], and an analysis of the mixed unitary rank of channels [16]. We also mention earlier work on the class from different perspectives [3, 17, 22].

Below we present a theorem that provides a connection between mixed unitary channels and nullspaces of entanglement breaking channels; first however we need the following result, which may be found as theorem 1 in [16], but we will provide a short proof here for completeness.

Lemma 4.2. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be a channel with canonical complement ${{\Phi}}^{\mathrm{C}}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{d}\left(\mathbb{C}\right)$. Then Φ is mixed unitary implemented with r unitaries if and only if there exists an isometry $W:{\mathbb{C}}^{d}\to {\mathbb{C}}^{r}$ such that, for all $X\in {M}_{n}\left(\mathbb{C}\right)$ with Tr(X) = 0, the matrix WΦ^C(X)W* has all of its diagonal entries equal to 0.

Proof. Suppose first that Φ is mixed unitary; then there exist unitaries ${\left\{{U}_{i}\right\}}_{i=1}^{r}$ and probabilities ${\left\{{p}_{i}\right\}}_{i=1}^{r}$ such that the (i, j) entry of the matrix Φ^C(X) is equal to $\mathrm{Tr}\left(\sqrt{{p}_{i}{p}_{j}}{U}_{j}^{{\ast}}{U}_{i}X\right)$. Setting i = j we get p_i  Tr(X) and so for all traceless X, the diagonal entries of this matrix are 0.

For the converse, suppose an isometry $W={\left({w}_{ij}\right)}_{r{\times}d}$ exists with the property that each of the diagonal entries of WΦ^C(X)W* are zero for all traceless X. Define $\tilde {{V}_{i}}={\sum }_{j=1}^{d}{w}_{ij}{V}_{i}$ for 1 ⩽ i ⩽ d, where {V_i } are a set of Kraus operators for Φ. Then $\left\{\tilde {{V}_{i}}\right\}$ is also a set of Kraus operators for Φ as W is an isometry, and one can check the (i, j) entry satisfies ${\left(W{{\Phi}}^{\mathrm{C}}\left(X\right){W}^{{\ast}}\right)}_{ij}=\mathrm{Tr}\left({\tilde {{V}_{j}}}^{{\ast}}\tilde {{V}_{i}}X\right)$. In particular, we have $\mathrm{Tr}\left({\tilde {{V}_{i}}}^{{\ast}}\tilde {{V}_{i}}X\right)=0$ for all i, and for all traceless X. Hence we have ${\tilde {{V}_{i}}}^{{\ast}}\tilde {{V}_{i}}\in {\left\{I\right\}}^{{\perp }^{\perp }}$, and so ${\tilde {{V}_{i}}}^{{\ast}}\tilde {{V}_{i}}$ is a (non-zero) multiple of the identity: ${\tilde {{V}_{i}}}^{{\ast}}\tilde {{V}_{i}}={p}_{i}I$. Thus, ${U}_{i}{:=}\frac{1}{\sqrt{{p}_{i}}}\tilde {{V}_{i}}$ is unitary, and $\tilde {{V}_{i}}=\sqrt{{p}_{i}}{U}_{i}$. That the set ${\left\{{p}_{i}\right\}}_{i=1}^{r}$ forms a probability distribution follows from trace preservation of the original map:

$$\begin{equation*}\sum\limits _{i}{p}_{i}I=\left[{\tilde {{V}_{1}}}^{{\ast}}{\tilde {{V}_{2}}}^{{\ast}}\dots \right]{\left[\tilde {{V}_{1}}\tilde {{V}_{2}}\dots \right]}^{t}=\left[{V}_{1}^{{\ast}}\dots \enspace \right]{W}^{{\ast}}W{\left[{V}_{1}\dots \right]}^{t}=\sum\limits _{i}{V}_{i}^{{\ast}}{V}_{i}=I;\end{equation*}$$

the last equality using the fact that the {V_i } are Kraus operators for Φ. □

We use the term r -dimensional diagonal algebra to mean an algebra $\mathcal{A}\subseteq {M}_{r}\left(\mathbb{C}\right)$ that is unitarily equivalent to the (commutative) algebra of diagonal matrices Δ_r inside ${M}_{r}\left(\mathbb{C}\right)$. Note that any orthonormal basis ${\left\{{u}_{i}\right\}}_{i=1}^{r}$ for ${\mathbb{C}}^{r}$ generates an r-dimensional diagonal algebra via $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{{u}_{i}{u}_{i}^{{\ast}}\right\}\cong {{\Delta}}_{r}$. Given such an algebra $\mathcal{A}$ and a positive matrix P inside it, we can define a (positive) trace functional on $\mathcal{A}$ by Tr_P (D) := Tr(PD) for $D\in \mathcal{A}$. We can then consider completely positive maps ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to \mathcal{A}$ that are trace-preserving from the regular trace on matrices to this modified trace on $\mathcal{A}$; that is, Tr_P (Φ(X)) = Tr(X) for $X\in {M}_{n}\left(\mathbb{C}\right)$. We shall also call such maps channels when the algebra and modified trace is understood.

We now present our theorem on a mixed unitary channel test.

Theorem 4.3. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be a channel with canonical complement ${{\Phi}}^{\mathrm{C}}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{d}\left(\mathbb{C}\right)$. Then Φ is mixed unitary and implemented with r unitaries if and only if there is an entanglement breaking map $E:{M}_{d}\left(\mathbb{C}\right)\to {M}_{r}\left(\mathbb{C}\right)$ of Choi rank at most r, with Kraus operators ${\left\{{u}_{i}{w}_{i}^{{\ast}}\right\}}_{i=1}^{r}$ such that u_i ≠ 0 are orthogonal, E is a channel with respect to the diagonal algebra defined by the u_i , and E privatizes the range of Φ^C to a projection $Q\in {M}_{r}\left(\mathbb{C}\right)$; that is,

$$\begin{equation*}E\left({{\Phi}}^{\mathrm{C}}\left(X\right)\right)=\mathrm{Tr}\left(X\right)\enspace Q\quad \forall X\in {M}_{n}\left(\mathbb{C}\right).\end{equation*}$$

Proof. First suppose Φ is mixed unitary and implemented with r multiples of unitaries. By lemma 4.2 there must be an isometry $W:{\mathbb{C}}^{d}\to {\mathbb{C}}^{r}$ such that WΦ^C(X)W* has 0 on its diagonal when X is traceless. Let ${w}_{1},\dots ,{w}_{r}\in {\mathbb{C}}^{d}$ be the columns of W*; then ${w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left(X\right){w}_{i}=0$ for all traceless X and all i. Also the condition that W is an isometry may be phrased as ${I}_{d}={W}^{{\ast}}W={\sum }_{i=1}^{r}{w}_{i}{w}_{i}^{{\ast}}$.

Define ${p}_{i}=\frac{1}{n}{w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({I}_{n}\right){w}_{i}$. As Φ^C is trace preserving and W an isometry, we have that

$$\begin{equation*}\sum\limits _{i=1}^{r}{p}_{i}=\frac{1}{n}\sum\limits _{i=1}^{r}{w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({I}_{n}\right){w}_{i}=\frac{1}{n}\sum\limits _{i=1}^{r}\mathrm{Tr}\left({{\Phi}}^{\mathrm{C}}\left({I}_{n}\right){w}_{i}{w}_{i}^{{\ast}}\right)=\frac{1}{n}\mathrm{Tr}\left({I}_{n}\right)=1.\end{equation*}$$

Let ${\left\{\tilde {{u}_{i}}\right\}}_{i=1}^{r}$ be any orthonormal basis for ${\mathbb{C}}^{r}$. Let s ⩽ r be the number of non-zero p_i and re-order so that p₁, ..., p_s are non-zero. Then define an orthogonal basis for ${\mathbb{C}}^{r}$ by ${u}_{i}={\sqrt{{p}_{i}}}^{-1}\tilde {{u}_{i}}$ for 1 ⩽ i ⩽ s and ${u}_{i}=\tilde {{u}_{i}}$ otherwise. Next define the entanglement breaking map E to have Kraus operators ${\left\{{u}_{i}{w}_{i}^{{\ast}}:{\mathbb{C}}^{d}\to {\mathbb{C}}^{r}\right\}}_{i=1}^{r}$, which has Choi rank at most r and range contained in the diagonal algebra $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{u}_{i}{u}_{i}^{{\ast}}\right\}}_{i=1}^{r}\cong {{\Delta}}_{r}$.

Now, for any $X\in {M}_{n}\left(\mathbb{C}\right)$, write X = n⁻¹ Tr(X)I_n + X₀ where X₀ is traceless, and observe

$$\begin{equation*}E\left({{\Phi}}^{\mathrm{C}}\left({X}_{0}\right)\right)=\sum\limits _{i=1}^{r}{u}_{i}\left({w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({X}_{0}\right){w}_{i}\right){u}_{i}^{{\ast}}=0.\end{equation*}$$

That is, E annihilates the traceless part of Range(Φ^C). Thus it remains to see what E does to Φ^C(I_n ). From the definition of the p_i we have,

$$\begin{equation*}E\left({{\Phi}}^{\mathrm{C}}\left({I}_{n}\right)\right)=\sum\limits _{i=1}^{s}\frac{1}{{p}_{i}}\tilde {{u}_{i}}\left({w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({I}_{n}\right){w}_{i}\right){\tilde {{u}_{i}}}^{{\ast}}=n\sum\limits _{i=1}^{s}\frac{{p}_{i}}{{p}_{i}}\tilde {{u}_{i}}{\tilde {{u}_{i}}}^{{\ast}}=n\enspace Q,\end{equation*}$$

where Q is a rank-s projection in ${M}_{r}\left(\mathbb{C}\right)$.

Regarding trace-preservation, let V be the unitary that implements the unitary equivalence of $\mathcal{A}$ with Δ_r such that ${V}^{{\ast}}{{\Delta}}_{r}V=\mathcal{A}$ and

$$\begin{equation*}P{:=}{V}^{{\ast}}\enspace \mathrm{diag}\left({p}_{1},\dots ,{p}_{s},1,\dots ,1\right)V=\sum\limits _{i=1}^{s}{p}_{i}\tilde {{u}_{i}}{\tilde {{u}_{i}}}^{{\ast}}+\sum\limits _{i{ >}s}\tilde {{u}_{i}}{\tilde {{u}_{i}}}^{{\ast}},\end{equation*}$$

which is a positive definite matrix in $\mathcal{A}$. Hence we may define a trace on $\mathcal{A}$ by Tr_P (D) = Tr(PD) for $D\in \mathcal{A}$. Then E is trace-preserving with respect to the regular trace on ${M}_{d}\left(\mathbb{C}\right)$ and this trace on $\mathcal{A}$. This follows because for all $X\in {M}_{d}\left(\mathbb{C}\right)$, we have

$$\begin{align*}\hfill {\mathrm{Tr}}_{P}\left(E\left(X\right)\right)& =\sum\limits _{i=1}^{r}\mathrm{Tr}\left(P{u}_{i}{w}_{i}^{{\ast}}X{w}_{i}{u}_{i}^{{\ast}}\right)\hfill \\ \hfill & =\sum\limits _{i=1}^{s}{p}_{i}\enspace \mathrm{Tr}\left({u}_{i}{w}_{i}^{{\ast}}X{w}_{i}{u}_{i}^{{\ast}}\right)+\sum\limits _{i{ >}s}\mathrm{Tr}\left(\tilde {{u}_{i}}{w}_{i}^{{\ast}}X{w}_{i}{\tilde {{u}_{i}}}^{{\ast}}\right)\hfill \\ \hfill & =\sum\limits _{i=1}^{s}\frac{{p}_{i}}{{p}_{i}}\enspace \mathrm{Tr}\left({w}_{i}^{{\ast}}X{w}_{i}\right)+\sum\limits _{i{ >}s}\mathrm{Tr}\left({w}_{i}^{{\ast}}X{w}_{i}\right)\hfill \\ \hfill & =\mathrm{Tr}\left(\left(\sum\limits _{i=1}^{r}{w}_{i}{w}_{i}^{{\ast}}\right)X\right)=\mathrm{Tr}\left(X\right).\hfill \end{align*}$$

For the other direction, suppose $E:{M}_{d}\left(\mathbb{C}\right)\to {M}_{r}\left(\mathbb{C}\right)$ exists with the listed properties, and Kraus operators ${\left\{{u}_{i}{w}_{i}^{{\ast}}\right\}}_{i=1}^{r}$ such that {u_i } form an orthogonal basis for ${\mathbb{C}}^{r}$.

Then, for any traceless X₀, we have that

$$\begin{equation*}0=E\left({{\Phi}}^{\mathrm{C}}\left({X}_{0}\right)\right)=\sum\limits _{i=1}^{r}\left({w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({X}_{0}\right){w}_{i}\right){u}_{i}{u}_{i}^{{\ast}}.\end{equation*}$$

Since {u_i } are orthogonal, the rank one operators ${u}_{i}{u}_{i}^{{\ast}}$ are linearly independent and so ${w}_{i}^{{\ast}}{{\Phi}}^{\mathrm{C}}\left({X}_{0}\right){w}_{i}=0$ for all i.

Finally, since E is trace-preserving between the regular trace on ${M}_{d}\left(\mathbb{C}\right)$ and Tr_P on $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{u}_{i}{u}_{i}^{{\ast}}\right\}}_{i=1}^{r}$, where $P\in \mathcal{A}$ is positive, using the equation Tr(X) = Tr(PE(X)) for all $X\in {M}_{d}\left(\mathbb{C}\right)$ and rearranging inside the trace, yields:

$$\begin{equation*}\sum\limits _{i=1}^{r}{w}_{i}{u}_{i}^{{\ast}}P{u}_{i}{w}_{i}^{{\ast}}={I}_{d}.\end{equation*}$$

Hence, the matrix W* with columns ${\Vert}\sqrt{P}{u}_{i}{\Vert}{w}_{i}$, for 1 ⩽ i ⩽ r, is a co-isometry from ${\mathbb{C}}^{r}$ into ${\mathbb{C}}^{d}$, with the property that WΦ^C(X₀)W* has zeroes on the diagonal for all $X\in {M}_{n}\left(\mathbb{C}\right)$. By lemma 4.2, Φ must be mixed unitary implemented with (at most) r unitaries, and this completes the proof. □

Remark 4.4. We note an interesting connection between this result and a result from [43], where it was proved that any quantum channel (in the regular trace-preserving sense) whose range is contained in a diagonal algebra must be entanglement breaking.

The following pair of examples illustrate the mechanics of the theorem construction and the test it provides in special cases of interest.

Example 4.5. Let ${{\Phi}}_{\mathrm{C}\mathrm{D}}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be the completely depolarizing map, recall as characterized by ${{\Phi}}_{\mathrm{C}\mathrm{D}}\left(X\right)=\frac{\mathrm{Tr}\left(X\right)}{n}{I}_{n}$ for all $X\in {M}_{n}\left(\mathbb{C}\right)$. One set of Kraus operators for this map is ${\left\{\frac{1}{\sqrt{n}}{E}_{ij}\right\}}_{i,j=1}^{n}$, and hence the canonical complement is given by

$$\begin{equation*}{{\Phi}}_{\mathrm{C}\mathrm{D}}^{\mathrm{C}}\left(X\right)=\sum\limits _{i,j,r,s=1}^{n}\mathrm{Tr}\left({E}_{sr}{E}_{ij}X\right){E}_{ir}\otimes {E}_{js};\end{equation*}$$

and thus ${{\Phi}}_{\mathrm{C}\mathrm{D}}^{\mathrm{C}}\left(X\right)={I}_{n}\otimes X$.

Let ${\left\{{U}_{i}\right\}}_{i=1}^{n}$ be any set of mutually orthogonal unitaries in the trace inner product on ${\mathbb{C}}^{n}$; for example the Weyl unitaries W_ij := Xⁱ Z^j where X is the cyclic shift and Z is diagonal with diagonal entry Z_ii = ωⁱ , where ω is a primitive nth root of unity.

Let u_i = vec(U_i ), the vector obtained by stacking the columns of U_i into a column vector. It is well known that vec(XU_i ) = (I_n ⊗ X)u_i and hence

$$\begin{equation*}\mathrm{Tr}\left(X\right)=\mathrm{Tr}\left({U}_{i}^{{\ast}}X{U}_{i}\right)=\langle {u}_{i},\left({I}_{n}\otimes X\right){u}_{i}\rangle .\end{equation*}$$

As the u_i are mutually orthogonal, the matrix V with columns $\frac{1}{\sqrt{n}}{u}_{i}$ is a unitary, and satisfies

$$\begin{equation*}{\left[{V}^{{\ast}}{{\Phi}}_{\mathrm{C}\mathrm{D}}^{\mathrm{C}}\left(X\right)V\right]}_{ii}=\frac{1}{n}\langle {u}_{i},\left({I}_{n}\otimes X\right){u}_{i}\rangle =0\end{equation*}$$

whenever Tr(X) = 0 and so Φ_CD must be mixed unitary. Indeed, one can verify directly that the map Φ_CD is implemented with Kraus operators given by any maximal set of orthogonal unitaries, evenly scaled for trace preservation.

Hence, if we form the entanglement breaking channel $E:{M}_{{n}^{2}}\left(\mathbb{C}\right)\to {M}_{{n}^{2}}\left(\mathbb{C}\right)$ to have Kraus operators ${\left\{\frac{1}{\sqrt{n}}{e}_{i}{u}_{i}^{{\ast}}\right\}}_{i=1}^{{n}^{2}}$, where e_i is the standard basis for ${\mathbb{C}}^{{n}^{2}}$, we see that

$$\begin{equation*}E\left({{\Phi}}_{\mathrm{C}\mathrm{D}}^{\mathrm{C}}\left(X\right)\right)=E\left({I}_{n}\otimes X\right)=\frac{1}{n}\sum\limits _{i=1}^{{n}^{2}}\langle {u}_{i},\left({I}_{n}\otimes X\right){u}_{i}\rangle {E}_{ii}=\frac{\mathrm{Tr}\left(X\right)}{n}{I}_{{n}^{2}}.\end{equation*}$$

Example 4.6. Consider the Werner–Holevo channel ${\Phi}:{M}_{3}\left(\mathbb{C}\right)\to {M}_{3}\left(\mathbb{C}\right)$ defined by ${\Phi}\left(X\right)=\frac{1}{2}\left(\mathrm{Tr}\left(X\right)I-{X}^{\mathrm{t}}\right)$, where X^t denotes the transpose of X. It is well known that this map is not mixed unitary (see [31]). We will use theorem 4.3 to detect this fact.

One can check that a set of Kraus operators for Φ are given by the following three matrices:

$$\begin{equation*}{K}_{1}=\left[\begin{matrix}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill \frac{-1}{\sqrt{2}}\hfill & \hfill 0\hfill \end{matrix}\right],\qquad {K}_{2}=\left[\begin{matrix}\hfill 0\hfill & \hfill 0\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{1}{\sqrt{2}}\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right],\qquad {K}_{3}=\left[\begin{matrix}\hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill 0\hfill \\ \hfill \frac{-1}{\sqrt{2}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right].\end{equation*}$$

Now it follows that in this case the complementary channel is Φ itself; that is, Φ^C = Φ as can be verified directly from the definition of Φ^C. Also the channel has Choi rank equal to 3, and so we have ${{\Phi}}^{\mathrm{C}}={\Phi}:{M}_{3}\left(\mathbb{C}\right)\to {M}_{3}\left(\mathbb{C}\right)$. Suppose Φ is mixed unitary with r unitaries. Then by the proof of theorem 4.3 we have an entanglement breaking map $E:{M}_{3}\left(\mathbb{C}\right)\to {M}_{r}\left(\mathbb{C}\right)$ of Choi rank at most r such that $E\left({{\Phi}}^{\mathrm{C}}\left(X\right)\right)=\frac{1}{r}\enspace \mathrm{Tr}\left(X\right){I}_{r}$. As the range of Φ = Φ^C is the whole matrix space ${M}_{3}\left(\mathbb{C}\right)$, the entanglement breaking map E is essentially the completely depolarizing map $X{\mapsto}\mathrm{Tr}\left(X\right)\frac{{I}_{r}}{r}$ from ${M}_{3}\left(\mathbb{C}\right)$ to ${M}_{r}\left(\mathbb{C}\right)$. However, we know that this map has Choi rank 3r, which gives a contradiction.

In this section, we build on the nullspace analyses above to derive constructions of algebras privatized by certain entanglement breaking channels. We first review some details of an important operator structure from operator theory [10], which in more recent years has also found a role in quantum information [11, 24, 29, 39, 40].

Definition 5.1. The multiplicative domain, ${\mathcal{M}}_{{\Phi}}$, of a completely positive map ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ is the *-subalgebra of ${M}_{n}\left(\mathbb{C}\right)$ given by:

$$\begin{equation*}\left\{A\in {M}_{n}\left(\mathbb{C}\right):{\Phi}\left(AX\right)={\Phi}\left(A\right){\Phi}\left(X\right);{\Phi}\left(XA\right)={\Phi}\left(X\right){\Phi}\left(A\right)\quad \forall \enspace X\in {M}_{n}\left(\mathbb{C}\right)\right\}.\end{equation*}$$

We note that for unital maps (Φ(I) = I), a projection P belongs to ${\mathcal{M}}_{{\Phi}}$ if and only if Φ(P) is a projection [39]. From [40], we also know that for any unital PPT map Φ (and in particular this applies to the dual Φ^† of any entanglement breaking channel), and any projection P in the multiplicative domain ${\mathcal{M}}_{{\Phi}}$, that Φ(X) = Φ(PXP) + Φ(QXQ) = Φ(P)Φ(X)Φ(P) + Φ(Q)Φ(X)Φ(Q) for all $X\in {M}_{n}\left(\mathbb{C}\right)$, where Q := I − P. It is also the case that Φ(P)Φ(X) = Φ(X)Φ(P). We can use this to show that if P, Q are orthogonal projections in the multiplicative domain, that any X for which PXQ = X must satisfy Φ(X) = 0, as

$$\begin{equation*}{\Phi}\left(X\right)={\Phi}\left(PXQ\right)={\Phi}\left(P\right){\Phi}\left(X\right){\Phi}\left(Q\right)={\Phi}\left(PQ\right){\Phi}\left(X\right)=0.\end{equation*}$$

The following structural result on the multiplicative domain of dual maps for entanglement breaking channels is used in our results below and may be of independent interest. We remind the reader that the dual of the completely positive map ${\Phi}\left(X\right)={\sum }_{i}{V}_{i}X{V}_{i}^{{\ast}}$ (denoted by Φ^†) is the completely positive (and unital when Φ is trace preserving) map given by ${{\Phi}}^{{\dagger}}\left(X\right)={\sum }_{i}{V}_{i}^{{\ast}}X{V}_{i}$.

Lemma 5.2. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be an entanglement breaking channel given by equation (2). Let ${\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$ be the multiplicative domain of Φ^†, and let ${\left\{{P}_{k}\right\}}_{k=1}^{r}\subseteq {\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$ be a set of mutually orthogonal projections summing to the identity. Then for all i there is a unique k such that v_i = P_k v_i .

Further let ${\mathcal{R}}_{k}\subseteq \left\{1,2,\dots ,d\right\}$ for 1 ⩽ k ⩽ r be the subsets determined by the partition generated by the P_k , and define ${\mathcal{W}}_{k}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{w}_{j}\right\}}_{j\in {\mathcal{R}}_{k}}$. Then ${\mathcal{W}}_{k}$ are mutually orthogonal subspaces and the projections Q_k onto ${\mathcal{W}}_{k}$ are a set of mutually orthogonal projections summing to the identity. Moreover, for all $X\in {M}_{n}\left(\mathbb{C}\right)$ and 1 ⩽ k ⩽ r we have

$$\begin{equation*}{\Phi}\left({Q}_{k}X\right)={P}_{k}{\Phi}\left(X\right)={\Phi}\left(X\right){P}_{k}={\Phi}\left(X{Q}_{k}\right),\end{equation*}$$

and so if X = Q_k XQ_l with k ≠ l, then Φ(X) = 0.

Proof. Since each ${P}_{k}\in {\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$, for k ≠ l we have

$$\begin{equation*}{{\Phi}}^{{\dagger}}\left({P}_{k}\right){{\Phi}}^{{\dagger}}\left({P}_{l}\right)={{\Phi}}^{{\dagger}}\left({P}_{k}{P}_{l}\right)=0.\end{equation*}$$

Also since ${{\Phi}}^{{\dagger}}\left(X\right)={\sum }_{i}{w}_{i}{v}_{i}^{{\ast}}X{v}_{i}{w}_{i}^{{\ast}}$, if we denote R_k = Φ^†(P_k ) then we have for k ≠ l,

$$\begin{equation*}0=\mathrm{Tr}\left({R}_{k}{R}_{l}\right)=\sum\limits _{i,j}\left({v}_{i}^{{\ast}}{P}_{k}{v}_{i}\right)\left({v}_{j}^{{\ast}}{P}_{l}{v}_{j}\right)\vert \langle {w}_{i},{w}_{j}\rangle {\vert }^{2}.\end{equation*}$$

So every term in the sum must be zero, and in particular when i = j, we have ${\Vert}{w}_{i}{{\Vert}}^{2}\left({v}_{i}^{{\ast}}{P}_{k}{v}_{i}\right)\left({v}_{j}^{{\ast}}{P}_{l}{v}_{j}\right)=0$. Hence for each i, ${v}_{i}^{{\ast}}{P}_{k}{v}_{i}{ >}0$ for at most one k. As ||v_i || = 1 and ∑_k P_k = I, we must have exactly one index 1 ⩽ k ⩽ r such that P_k v_i = v_i .

Let ${\mathcal{V}}_{k}=\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\left({P}_{k}\right)$, so that ${\mathbb{C}}^{n}={\oplus }_{i=1}^{r}{V}_{k}$ is an orthogonal direct sum decomposition of ${\mathbb{C}}^{n}$. Thus, the projections P_k impose a partition of {1, 2, ..., d} into subsets ${\mathcal{R}}_{k}$ such that ${\mathcal{V}}_{k}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{v}_{j}\right\}}_{j\in {\mathcal{R}}_{k}}$. (Note it may be the case that some ${\mathcal{R}}_{k}=\varnothing$.) Next we show that the same partition also induces an orthogonal direct sum structure on the ${\left\{{w}_{i}\right\}}_{i=1}^{d}$ vectors.

Since ${P}_{k}{v}_{j}={\chi }_{j\in {\mathcal{R}}_{k}}{v}_{j}$ where χ is the indicator function, we may write

$$\begin{equation*}{R}_{k}{:=}{{\Phi}}^{{\dagger}}\left({P}_{k}\right)=\sum\limits _{j\in {\mathcal{R}}_{k}}\left({v}_{j}^{{\ast}}{P}_{k}{v}_{j}\right){w}_{j}{w}_{j}^{{\ast}}=\sum\limits _{j\in {\mathcal{R}}_{k}}{w}_{j}{w}_{j}^{{\ast}}.\end{equation*}$$

As the R_k have mutually orthogonal ranges, for k ≠ l we have

$$\begin{equation*}0=\mathrm{Tr}\left({R}_{k}{R}_{l}\right)=\sum\limits _{i\in {\mathcal{R}}_{k},j\in {\mathcal{R}}_{l}}\vert \langle {w}_{i},{w}_{j}\rangle {\vert }^{2},\end{equation*}$$

so each term in the sum is zero, and it follows that ${\mathcal{W}}_{k}$ and ${\mathcal{W}}_{l}$ are orthogonal.

Hence, the projections {Q_k } have mutually orthogonal ranges. Further, the subspace spanned by their (projection) sum Q = ∑_k Q_k must be the identity as Q projects onto ${\cup }_{k=1}^{r}{\mathcal{W}}_{k}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}{\left\{{w}_{i}\right\}}_{i=1}^{d}={\mathbb{C}}^{n}$; the last equality following from $I={\sum }_{i}{w}_{i}{w}_{i}^{{\ast}}$.

Finally, for 1 ⩽ k ⩽ r we can compute:

$$\begin{align*}\hfill {\Phi}\left({Q}_{k}X\right)& =\sum\limits _{i=1}^{d}\left({w}_{i}^{{\ast}}{Q}_{k}X{w}_{i}\right){v}_{i}{v}_{i}^{{\ast}}\hfill \\ \hfill & =\sum\limits _{i\in {\mathcal{R}}_{k}}\left({w}_{i}^{{\ast}}X{w}_{i}\right){v}_{i}{v}_{i}^{{\ast}}\hfill \\ \hfill & =\sum\limits _{i=1}^{d}\left({w}_{i}^{{\ast}}X{w}_{i}\right){P}_{k}{v}_{i}{v}_{i}^{{\ast}}\hfill \\ \hfill & ={P}_{k}{\Phi}\left(X\right),\hfill \end{align*}$$

and the other equalities are proved in the same way. Specifically, these equalities imply for k ≠ l:

$$\begin{equation*}{\Phi}\left(X\right)={\Phi}\left({Q}_{k}X{Q}_{l}\right)={P}_{k}{\Phi}\left(X{Q}_{l}\right)={P}_{k}{P}_{l}{\Phi}\left(X\right)=0,\end{equation*}$$

and the result follows. □

Remark 5.3. From a nullspace perspective, note that in any basis which mutually diagonalizes all Q_k simultaneously, with block matrix structure corresponding to the division of ${\mathbb{C}}^{n}$ into direct summands ${\mathcal{W}}_{k}$, any matrix X that is supported entirely on the off-diagonal blocks of the decomposition is annihilated by Φ. In equation form, this says for all $X\in {M}_{n}\left(\mathbb{C}\right)$ that Φ(X) = ∑_j,k Φ(Q_j XQ_k ) = ∑_k Φ(Q_k XQ_k ) = ∑_k P_k Φ(X)P_k .

The completely depolarizing channel discussed above obviously privatizes the full algebra ${M}_{n}\left(\mathbb{C}\right)$, and in that case ${\mathcal{M}}_{{{\Phi}}^{{\dagger}}}=\mathbb{C}I$, and P₁ = I = Q₁. More generally, given that the vectors v_i , w_i which determine the rank-one form of an entanglement breaking channel can be arbitrary, up to the trace preservation condition being satisfied, it is reasonable to expect that generic channels from the class will not privatize any non-trivial algebra. Nevertheless, based on the analysis above, we finish by identifying two special classes of channels that do privatize algebras.

Theorem 5.4. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be an entanglement breaking channel, with operator-sum form as given in equation (2). Suppose ${\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$ contains a rank-one projection P = v v*. Then Φ privatizes the algebra $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{{w}_{i}{w}_{j}^{{\ast}}:P{v}_{i}={v}_{i},P{v}_{j}={v}_{j}\right\}$ to P; that is,

$$\begin{equation*}{\Phi}\left(A\right)=\mathrm{Tr}\left(A\right)P\quad \forall \enspace A\in \mathcal{A}.\end{equation*}$$

Proof. First note that $\mathcal{A}$ is indeed an *-algebra, even though it is only defined as linearly closed, as it is a self-adjoint operator space and closed under multiplication.

Given the Kraus operators ${\left\{{v}_{i}{w}_{i}^{{\ast}}\right\}}_{i=1}^{d}$ for Φ, suppose we have a (nonempty) subset ${\mathcal{R}}_{v}\subset \left\{1,\dots ,d\right\}$ and unit vector $v\in {\mathbb{C}}^{n}$ such that v = v_i for all $i\in {\mathcal{R}}_{v}$. Then $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{{w}_{i}{w}_{j}^{{\ast}}:i,j\in {\mathcal{R}}_{v}\right\}$. Put P = vv* and let Q be the projection onto $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{{w}_{i}:i\in {\mathcal{R}}_{v}\right\}$ as in the proof of lemma 5.2.

To complete the proof it is enough to show that ${\Phi}\left({w}_{i}{w}_{j}^{{\ast}}\right)=\mathrm{Tr}\left({w}_{i}{w}_{j}^{{\ast}}\right)P$ for any fixed $i,j\in {\mathcal{R}}_{v}$. This follows from multiple applications of lemma 5.2 in the following calculation, in which we take an arbitrary $X\in {M}_{n}\left(\mathbb{C}\right)$:

$$\begin{align*}\hfill \mathrm{Tr}\left({\Phi}\left({w}_{i}{w}_{j}^{{\ast}}\right)X\right)& =\mathrm{Tr}\left({\Phi}\left(Q{w}_{i}{w}_{j}^{{\ast}}Q\right)X\right)\hfill \\ \hfill & =\mathrm{Tr}\left(P{\Phi}\left({w}_{i}{w}_{j}^{{\ast}}\right)PX\right)\hfill \\ \hfill & =\left({v}^{{\ast}}Xv\right)\mathrm{Tr}\left(P{\Phi}\left({w}_{i}{w}_{j}^{{\ast}}\right)P\right)\hfill \\ \hfill & =\left({v}^{{\ast}}Xv\right)\mathrm{Tr}\left({\Phi}\left(Q{w}_{i}{w}_{j}^{{\ast}}Q\right)\right)\hfill \\ \hfill & =\left({v}^{{\ast}}Xv\right)\mathrm{Tr}\left({\Phi}\left({w}_{i}{w}_{j}^{{\ast}}\right)\right)\hfill \\ \hfill & =\left({v}^{{\ast}}Xv\right)\mathrm{Tr}\left({w}_{i}{w}_{j}^{{\ast}}\right)\hfill \\ \hfill & =\mathrm{Tr}\left(\mathrm{Tr}\left({w}_{i}{w}_{j}^{{\ast}}\right)PX\right),\hfill \end{align*}$$

and where the second last equality uses the trace preservation of Φ. As X was arbitrary, the result follows. □

Of course the algebra defined in the theorem could be trivial from a qubit encoding viewpoint—either {0} or having no matrix structure—but evidently there are many examples of entanglement breaking channels for which a non-trivial algebra and rank-one projection exist and satisfy the conditions of the theorem.

Example 5.5. Consider the physically described single qubit 'spontaneous emission' channel [35], ${\Phi}:{M}_{2}\left(\mathbb{C}\right)\to {M}_{2}\left(\mathbb{C}\right)$ given by ${\Phi}\left(\rho \right)={e}_{1}{e}_{1}^{{\ast}}$ for all density operators ρ; that is, Φ privatizes the entire algebra ${M}_{2}\left(\mathbb{C}\right)$ to $P={e}_{1}{e}_{1}^{{\ast}}$. Here we have two Kraus operators ${A}_{1}={E}_{11}={e}_{1}{e}_{1}^{{\ast}}$, ${A}_{2}={E}_{12}={e}_{1}{e}_{2}^{{\ast}}$, and so v₁ = e₁ = v₂, w₁ = e₁, w₂ = e₂ in the equation (2) form of the channel. The dual map satisfies ${{\Phi}}^{{\dagger}}\left(X\right)=\left({e}_{1}^{{\ast}}X{e}_{1}\right)I$ for all $X\in {M}_{2}\left(\mathbb{C}\right)$, and in particular the projection P belongs to ${\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$ (as it is mapped to a projection by Φ^†). Also we see the algebra from the theorem satisfies $\mathcal{A}={M}_{2}\left(\mathbb{C}\right)$ in this case. (In terms of the lemma notation, here ${P}_{1}={e}_{1}{e}_{1}^{{\ast}}$, ${P}_{2}={e}_{2}{e}_{2}^{{\ast}}$, Q₁ = I, Q₂ = 0.)

Similarly, higher dimensional versions of the spontaneous emission channel are covered by this result; ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$, with ${\Phi}\left(\rho \right)={e}_{1}{e}_{1}^{{\ast}}$, and Kraus operators ${A}_{k}={e}_{1}{e}_{k}^{{\ast}}$ for 1 ⩽ k ⩽ n. Here the algebra $\mathcal{A}$ is the full matrix algebra and it is privatized again to $P={e}_{1}{e}_{1}^{{\ast}}\in {\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$.

One can generalize this class of examples further, by considering entanglement breaking channels for which the vectors that determine the Kraus operators $\left\{{v}_{i}{w}_{i}^{{\ast}}\right\}$ have the property that an index subset ${\mathcal{R}}_{v}$ of the v_i satisfy v_i = v for some fixed vector v and all $i\in {\mathcal{R}}_{v}$. Then the algebra $\mathcal{A}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{{w}_{i}{w}_{j}^{{\ast}}:i,j\in {\mathcal{R}}_{v}\right\}$, which could have non-trivial structure depending on choice of the w_i , would be privatized to P = vv* by Φ. For instance, if $\vert {\mathcal{R}}_{v}\vert =k$ and $\left\{{w}_{i}:i\in {\mathcal{R}}_{v}\right\}$ is an orthogonal set of (non-zero) vectors, then $\mathcal{A}$ is unitarily equivalent to ${M}_{k}\left(\mathbb{C}\right)$ and would satisfy Φ(A) = Tr(A)P for all $A\in \mathcal{A}$.

We finish by identifying another class of entanglement breaking channels that privatize special types of matrix algebras.

Theorem 5.6. Let ${\Phi}:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ be an entanglement breaking channel. Suppose there are mutually orthogonal projections P_k , 1 ⩽ k ⩽ r, inside ${\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$ such that the projections Φ^†(P_k ) all have the same rank. Let $\mathcal{A}$ be any *-subalgebra of ${M}_{r}\left(\mathbb{C}\right)$ with constant diagonals. Then Φ privatizes an algebra *-isomorphic to $\mathcal{A}$.

Proof. Note first that the operators Q_k = Φ^†(P_k ) are indeed projections as each ${P}_{k}\in {\mathcal{M}}_{{{\Phi}}^{{\dagger}}}$. Let s = rank(Q_k ) and for each k put ${\mathcal{W}}_{k}={Q}_{k}{\mathbb{C}}^{n}$. For each pair 1 ⩽ k, l ⩽ r, let V_kl be a partial isometry on ${\mathbb{C}}^{n}$ with initial projection ${V}_{kl}^{{\ast}}{V}_{kl}={Q}_{l}$ and final projection ${V}_{kl}{V}_{kl}^{{\ast}}={Q}_{k}$. The V_ij set up an r × r block matrix picture for operators with domain and range supported on the range of the projection $Q={\sum }_{k=1}^{r}{Q}_{k}$.

We can then define an *-isomorphism ${\Psi}:{M}_{r}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$ by ${\Psi}\left(A\right)={\sum }_{k,l=1}^{r}{a}_{kl}{V}_{kl}$ for matrices A = (a_kl ). Note the image of Ψ inside ${M}_{n}\left(\mathbb{C}\right)$ is unitarily equivalent to ${M}_{r}\left(\mathbb{C}\right)\otimes {I}_{s}$. Moreover, as Ψ(A) = ∑_k,l a_kl Q_k V_kl Q_l , by lemma 5.2 we have

$$\begin{equation*}{\Phi}\left({\Psi}\left(A\right)\right)=\sum\limits _{k=1}^{r}{\Phi}\left({Q}_{k}{\Psi}\left(A\right){Q}_{k}\right)=\sum\limits _{k=1}^{r}{a}_{kk}{\Phi}\left({Q}_{k}\right)\end{equation*}$$

and Φ(Q_k ) = P_k Φ(I) = Φ(I)P_k , so that

$$\begin{equation*}{\Phi}\left({\Psi}\left(A\right)\right)=\left(\sum\limits _{k=1}^{r}{a}_{kk}{P}_{k}\right){\Phi}\left(I\right)={\Phi}\left(I\right)\left(\sum\limits _{k=1}^{r}{a}_{kk}{P}_{k}\right).\end{equation*}$$

Now consider an *-subalgebra $\mathcal{A}$ of ${M}_{r}\left(\mathbb{C}\right)$ with constant diagonals, ${a}_{kk}=\frac{\mathrm{Tr}\left(A\right)}{r}$ for all 1 ⩽ k ⩽ r and $A=\left({a}_{kl}\right)\in \mathcal{A}$. Let P = ∑_k P_k . Then from the above calculation we have for all $A\in \mathcal{A}$,

$$\begin{equation*}{\Phi}\left({\Psi}\left(A\right)\right)=\sum\limits _{k=1}^{r}\frac{1}{r}\enspace \mathrm{Tr}\left(A\right){P}_{k}{\Phi}\left(I\right)=\frac{\mathrm{Tr}\left(A\right)}{r}\left(P{\Phi}\left(I\right)\right)=\frac{\mathrm{Tr}\left(A\right)}{r}\left({\Phi}\left(I\right)P\right).\end{equation*}$$

Hence Φ privatizes the algebra ${\Psi}\left(\mathcal{A}\right)$ and the result follows. □

Remark 5.7. Algebras satisfying the condition of having all diagonal entries the same are plentiful and may be generated by taking any partition of the integer $r={\sum }_{k=1}^{p}{m}_{k}$, and factoring each part in the partition m_k = i_k j_k ; then there is a unitary U such that, after conjugation by U, the algebra ${\oplus }_{k=1}^{p}{I}_{\mathrm{max}{i}_{k},{j}_{k}}\otimes {M}_{\mathrm{min}{i}_{k},{j}_{k}}\left(\mathbb{C}\right)$ will have this form. The construction of such a U is somewhat intricate; see [[37], theorem 3.2.4] for details. In particular, for any expression of r as a sum of squares, $r={\sum }_{i=1}^{p}{i}_{k}^{2}$, we find that there is a unitary so that $U\left({\oplus }_{k=1}^{p}{I}_{{i}_{k}}\otimes {M}_{{i}_{k}}\left(\mathbb{C}\right)\right){U}^{{\ast}}$ has the property we seek.

Example 5.8. For an explicit example of a subclass of entanglement breaking channels that satisfy the conditions of the theorem, take a positive integer n with factors n = rs.

For each 1 ⩽ k ⩽ r, choose an orthonormal set of vectors $\left\{{w}_{i,k}:1{\leqslant}i{\leqslant}s\right\}\subseteq {\mathbb{C}}^{n}$ and let Q_k be the projection onto the (s-dimensional) subspace they span. Then, I = ∑_k Q_k and ${Q}_{k}={\sum }_{i=1}^{s}{w}_{i,k}{w}_{i,k}^{{\ast}}$. Next, for each 1 ⩽ k ⩽ r choose sets of non-zero (but not necessarily orthogonal) vectors $\left\{{v}_{i,k}:1{\leqslant}i{\leqslant}s\right\}\subseteq {\mathbb{C}}^{n}$, subject to the constraint ${v}_{i,k}^{{\ast}}{v}_{j,l}=0$ whenever k ≠ l. Let P_k be the projection onto the subspace spanned by ${\left\{{v}_{i,k}\right\}}_{i}$.

Let Φ be the entanglement breaking channel with Kraus operators

$$\begin{equation*}\left\{{v}_{i,k}{w}_{i,k}^{{\ast}}:\quad 1{\leqslant}i{\leqslant}s,\enspace 1{\leqslant}k{\leqslant}r\right\}.\end{equation*}$$

Then P₁, ..., P_r are a family of mutually orthogonal projections that belong to the multiplicative domain ${\mathcal{M}}_{{\Phi}}^{{\dagger}}$, as they are mapped to projections, in fact Φ^†(P_k ) = Q_k . Also recall each Q_k is rank-s. Thus, theorem 5.6 applies, and Φ privatizes a family of algebras that are isomorphic to subalgebras of ${M}_{r}\left(\mathbb{C}\right)$ with constant diagonals, as per the construction of the proof.

DWK was partly supported by NSERC and a University Research Chair at Guelph. RP was partly supported by NSERC. MR is partially supported by Research Initiation Grant at the BITS-Pilani Goa Campus.

No new data were created or analysed in this study.