On the Mixed-Unitary Rank of Quantum Channels

Abstract

In the theory of quantum information, the mixed-unitary quantum channels, for any positive integer dimension n, are those linear maps that can be expressed as a convex combination of conjugations by $n\times n$ complex unitary matrices. We consider the mixed-unitary rank of any such channel, which is the minimum number of distinct unitary conjugations required for an expression of this form. We identify several new relationships between the mixed-unitary rank N and the Choi rank r of mixed-unitary channels, the Choi rank being equal to the minimum number of nonzero terms required for a Kraus representation of that channel. Most notably, we prove that the inequality $N\le r^2-r+1$ is satisfied for every mixed-unitary channel (as is the equality $N=2$ when $r=2$), and we exhibit the first known examples of mixed-unitary channels for which $N>r$. Specifically, we prove that there exist mixed-unitary channels having Choi rank $d+1$ and mixed-unitary rank 2d for infinitely many positive integers d, including every prime power d. We also examine the mixed-unitary ranks of the mixed-unitary Werner–Holevo channels.

A. A characterization of Schur channels

Two channels $\Phi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ and $\Psi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ are said to be unitarily equivalent if there exist unitary matrices $U,V\in \mathcal {U}_n$ satisfying

$$\begin{aligned} \Psi (X) = U\Phi (VXV^*)U^* \end{aligned}$$

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for each $X\in \mathcal {M}_n$. In this appendix we provide necessary and sufficient conditions that characterize when a channel is unitarily equivalent to a Schur channel in terms of the channel’s operator system. We conclude from this characterization that the channels in Examples 10 and 11 are not equivalent to Schur channels.

It is known that a channel is a Schur map if and only if every Kraus representation for the channel consists of only diagonal matrices. The following lemma provides another useful characterization for Schur channels.

Lemma 24

A channel $\Phi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ is a Schur map if and only if $\Phi (D)=D$ holds for each diagonal matrix $D\in \mathcal {M}_n$.

Proof

If $\Phi (D)=D$ holds for each diagonal matrix $D\in \mathcal {M}_n$, then each Kraus matrix of $\Phi$ must commute with each diagonal matrix, and thus each Kraus matrix must itself be diagonal. On the other hand, if $\Phi$ is a Schur map then there is a correlation matrix $C\in \mathcal {M}_n$ satisfying $\Phi (X) = C\odot X$ for each $X\in \mathcal {M}_n$. As each diagonal entry of C must be equal to one, it holds that $\Phi (D) = D$ for each diagonal matrix $D\in \mathcal {M}_n$. $\quad \square$

We now provide a necessary and sufficient condition for characterizing when a map is unitarily equivalent to a Schur map in terms of the operator system of the channel.

Theorem 25

Let $\Phi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ be a channel. The following statements are equivalent.

• 1. The channel $\Phi$ is unitarily equivalent to a Schur map.

• 2. The operator system $\mathcal {S}_\Phi$ is a commuting family of matrices.

Proof

Let $A_1,\dots ,A_N\in \mathcal {M}_n$ be linear matrices satisfying

$$\begin{aligned} \Phi (X) = \sum _{k=1}^N A_kXA_k^* \end{aligned}$$

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for each $X\in \mathcal {M}_n$. First suppose that $\Phi$ is unitarily equivalent to a Schur map. There exist unitary matrices $U,V\in \mathcal {U}_n$ such that the channel $\Psi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ defined as $\Psi (X) = U\Phi (VXV^*)U^*$ is a Schur map. The channel $\Psi$ has a Kraus representation of the form

$$\begin{aligned} \Psi (X) = \sum _{j=1}^N (UA_jV)X(UA_jV)^* \end{aligned}$$

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and thus $UA_kV$ is a diagonal matrix for each $k\in \{1,\dots ,N\}$. Moreover, each of the matrices in the collection

$$\begin{aligned} V^*\mathcal {S}_\Phi V = {\text {span}}\{V^*A_j^*U^*UA_kV\, :\, j,k\in \{1,\dots ,N\}\} \end{aligned}$$

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is also diagonal. It follows that $\mathcal {S}_\Phi$ is a commuting family of normal matrices in $\mathcal {M}_n$.

For the other direction, suppose that $\mathcal {S}_\Phi$ is a commuting family. As $\mathcal {S}_\Phi$ is self-adjoint, each matrix in $\mathcal {S}_\Phi$ is also normal. There exists a unitary matrix $V\in \mathcal {U}_n$ such that $V^*A_j^*A_kV$ is a diagonal matrix for each pair of indices $j,k\in \{1,\dots ,N\}$. For any two diagonal matrices $D_0,D_1\in \mathcal {M}_n$, one has that

$$\begin{aligned} \Phi (VD_0V^*)\Phi (VD_1V^*)&= \sum _{j,k=1}^N A_jVD_0(V^*A_j^*A_kV)D_1V^*A_k^*\nonumber \\&= \sum _{j,k=1}^N A_jVD_0D_1(V^*A_j^*A_kV)V^*A_k^*\nonumber \\&= \Phi (VD_0D_1V^*)\Phi (\mathbbm {1}_n) \end{aligned}$$

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as each of the matrices in $\{V^*A_j^*A_kV\,:j,k\in \{1,\dots ,N\}\}$ is diagonal and commutes with the diagonal matrices $D_0$ and $D_1$. Define matrices $P_1,\dots ,P_n\in \mathcal {M}_n$ as

$$\begin{aligned} P_k = \Phi (VE_{k,k}V^*) \end{aligned}$$

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for each $k\in \{1,\dots ,n\}$. For indices $j,k\in \{1,\dots ,n\}$ with $j\ne k$, one has that

$$\begin{aligned} P_jP_k = \Phi (VE_{j,j}E_{k,k}V^*)\Phi (\mathbbm {1}_n) = 0 \end{aligned}$$

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as $E_{j,j}E_{k,k}=0$. Moreover, it holds that ${\text {Tr}}(P_k)={\text {Tr}}(E_{k,k})=1$ and that $P_k$ is positive for each $k\in \{1,\dots ,n\}$ as $\Phi$ is a quantum channel. The collection $\{P_1,\dots ,P_n\}\subset \mathcal {M}_n$ is therefore an orthogonal set of positive matrices each with trace equal to 1. Hence there must exist a unitary matrix $U\in \mathcal {U}_n$ such that $UP_jU^* = E_{j,j}$ for each $j\in \{1,\dots ,n\}$. Define a channel $\Psi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ as

$$\begin{aligned} \Psi (X) = U\Phi (VXV^*)U^* \end{aligned}$$

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for each $X\in \mathcal {M}_n$. From the observations above, one finds that $\Psi (E_{k,k})=E_{k,k}$ for each $k\in \{1,\dots ,n\}$, and thus $\Psi (D)=D$ for each diagonal matrix $D\in \mathcal {M}_n$. It follows that $\Psi$ is a Schur map by Lemma 24. This completes the proof. $\quad \square$

Remark

Every unital quantum channel with Choi rank at most 2 is unitarily equivalent to a Schur map. This fact was proven in [LS93], but we remark that another proof of this fact can be found by making use of Theorem 25. Indeed, let $\Phi :\mathcal {M}_n\rightarrow \mathcal {M}_n$ be a unital quantum channel for some positive integer n such that ${\text {rank}}(J(\Phi ))\le 2$. There exist matrices $A_0,A_1\in \mathcal {M}_n$ such that

$$\begin{aligned} \Phi (X) = A_0XA_0^* + A_1XA_1^* \end{aligned}$$

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is a Kraus representation of $\Phi$. As $\Phi$ is unital and trace preserving, these matrices must satisfy $A_0^*A_0 + A_1^*A_1 = \mathbbm {1}_n$ and $A_0 A_0^* + A_1A_1^*=\mathbbm {1}_n$. The operator system of $\Phi$ may be given by $\mathcal {S}_\Phi ={\text {span}}\{A_0^*A_0,A_0^*A_1,A_1^*A_0,A_1^*A_1\}$, and it is straightforward to verify that each of these matrices commute with one another:

$$\begin{aligned} (A_0^*A_0) (A_0^*A_1)= & {} A_0^*(\mathbbm {1}-A_1A_1^*)A_1 = A_0^*A_1(\mathbbm {1}-A_1^*A_1) = (A_0^*A_1)(A_0^*A_0),\nonumber \\ (A_0^*A_0) (A_1^*A_0)= & {} (\mathbbm {1}-A_1^*A_1)A_1^*A_0 = A_1^*(1-A_1A_1^*)A_0 = (A_1^*A_0)(A_0^*A_0),\nonumber \\ (A_1^*A_1) (A_0^*A_1)= & {} (\mathbbm {1}-A_0^*A_0)A_0^*A_1 = A_0^*(\mathbbm {1}-A_0A_0^*)A_1 = A_0^*A_1A_1^*A_1,\nonumber \\ (A_1^*A_1) (A_1^*A_0)= & {} A_1^*(\mathbbm {1}-A_0A_0^*)A_0 = A_1^*A_0(\mathbbm {1}-A_0^*A_0) = A_1^*A_0A_1^*A_1,\nonumber \\ (A_0^*A_0) (A_1^*A_1)= & {} (\mathbbm {1}-A_1^*A_1)(\mathbbm {1}-A_0^*A_0) = \mathbbm {1}- A_0^*A_0 - A_1^*A_1 +A_1^*A_1A_0^*A_0\nonumber \\= & {} (A_1^*A_1)(A_0^*A_0),\nonumber \\ (A_0^*A_1) (A_1^*A_0)= & {} A_0^*(\mathbbm {1}-A_0A_0^*)A_0 = A_0^*A_0(\mathbbm {1}-A_0^*A_0) = (\mathbbm {1}-A_1^*A_1)A_1^*A_1\nonumber \\= & {} A_1^*(\mathbbm {1}-A_1A_1^*)A_1 = (A_1^*A_0)(A_0^*A_1). \end{aligned}$$

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