Operator algebra generalization of a theorem of Watrous and mixed unitary quantum channels

Given a positive integer $d \unicode{x2A7E} 1$, we let M_d denote the set of d × d complex matrices. The matrix units E_ij , for $1 \unicode{x2A7D} i,j \unicode{x2A7D} d$, are the elements of M_d with a 1 in the $i,j$ entry and 0's elsewhere. The (Hilbert-Schmidt) trace inner product on M_d is given by $\langle A,B \rangle = \mathrm{Tr} (B^* A)$. The tensor product algebra $M_d \otimes M_d$ is naturally identified with $M_{d^2}$ and has matrix units $E_{ij}\otimes E_{kl}$. We will make use of the linear map $\mathrm{vec}:M_d \rightarrow \mathbb{C}^d \otimes \mathbb{C}^d$ defined by $\mathrm{vec}(E_{ij}) = e_i \otimes e_j$, where $\{e_1, \ldots ,e_{d} \}$ is the standard basis for $\mathbb{C}^d$.

We will be interested in completely positive maps $\Phi : M_d \rightarrow M_d$ [29], which can always be represented in operator-sum form $\Phi(X) = \sum_i K_i X K_i^*$ for some set of 'Kraus' operators $K_i \in M_d$ [19]. Each map defines a dual map via the trace inner product, wherein the roles of the operators K_i and $K_i^*$ are reversed in the operator-sum form. The map Φ is unital if $\Phi(I) = I$, where I is the identity matrix. If it is trace-preserving, which occurs exactly when $\sum_i K_i^* K_i = I$, then the map is called a quantum channel [16, 27, 37]. The class of unital (quantum) channels are pervasive in quantum information, and we define an important subclass below. Note that a channel is unital if and only its dual map is a unital channel as well.

The 'Choi matrix' [8] for Φ is the matrix $J(\Phi) \in M_d \otimes M_d$ given by,

$$\begin{align} J\left(\Phi\right) = \sum_{i,j = 1}^d E_{ij}\otimes \Phi\left(E_{ij}\right) . \end{align} \tag{ 1 }$$

It is a positive semi-definite matrix if and only if Φ is a completely positive map. The map $J(\cdot)$ is linear, and we note that $J(\Phi) = \sum_i \mathrm{vec}(K_i) \mathrm{vec}(K_i)^*$ when the K_i are Kraus operators for Φ [37].

In this paper, by an operator algebra we will mean a von Neumann algebra (or C$^*$-algebra) on finite-dimensional Hilbert space, which, up to unitarily equivalence [9], is a set of matrices contained inside some M_d of the form:

$$\begin{align} \mathcal{A} = \oplus_{k} \left(I_{m_k} \otimes M_{n_k}\right), \end{align} \tag{ 2 }$$

for some unique choice of positive integers $m_k, n_k$. The algebras we consider will typically be the fixed point sets of unital channels, and so necessarily will be unital ($I\in \mathcal{A}$), which means that $\sum_k m_k n_k = d$. The commutant $\mathcal{A}^{\prime}$ of $\mathcal{A}$, which is the set of all matrices in M_d that commute with every element of $\mathcal{A}$, has a corresponding form up to unitary equivalence given by $\mathcal{A}^{\prime} = \oplus_k (M_{m_k} \otimes I_{n_k})$.

Given an algebra $\mathcal{A} \subseteq M_d$, we can consider conditional expectations onto the algebra, which are maps $\mathcal{E} : M_d \rightarrow \mathcal{A}$ such that: (1) $\mathcal{E}(A) = A$ for all $A\in \mathcal{A}$; (2) $\mathcal{E}( A_1 X A_2) = A_1 \mathcal{E}(X) A_2$ for all $A_1,A_2\in \mathcal{A}$ and $X\in M_d$; and, (3) if $X\in M_d$ is positive semi-definite, then so is $\mathcal{E}(X)$. Every conditional expectation of M_d onto $\mathcal{A}$ is completely positive (and is a unital map when the algebra is unital), and amongst all possible conditional expectations onto $\mathcal{A}$, there is a unique map that is also trace-preserving [30] (in fact, it is exactly the orthogonal projection of M_d onto $\mathcal{A}$ in the trace inner product). So given a unital algebra $\mathcal{A} \subseteq M_d$, we shall denote the trace preserving conditional expectation onto $\mathcal{A}$ by $\mathcal{E}_{\mathcal{A}} : M_d \rightarrow \mathcal{A}$.

The fixed point set $\mathrm{Fix}(\Phi) = \{X \in M_d \, | \, \Phi(X) = X \}$ will also play a key role in our analysis. For a unital map Φ, it is easily seen that $\mathrm{Fix}(\Phi)$ contains the commutant of the Kraus operators, and further, for a unital channel these two sets coincide $\mathrm{Fix}(\Phi) = \{K_i \}^{\prime}$, and this statement is true for any choice of Kraus operators for the channel [20]. In particular, this means the fixed point set, which in general is just an operator subspace, in the case of unital channels is an operator algebra. (Note that this really is a feature of unital channels, as even the fixed point set of a unital completely positive map need not be an algebra [5, 20].)

We shall focus on the following class of unital channels, which are important in several areas of quantum information [3, 11, 12, 17, 21, 24].

Definition 1. A completely positive linear map $\Phi: M_d \rightarrow M_d$ is called a mixed unitary channel if there exists a set of d × d unitary matrices $\{U_i\}_{i = 1}^r$ and a set of nonnegative numbers $\{\lambda_i\}_{i = 1}^r$ with $\sum_{i = 1}^r \lambda_i = 1$ such that $\Phi(X) = \sum_{i = 1}^r \lambda_i U_iXU_i^*$.

It is known that every single-qubit unital channel is mixed unitary, but this is not the case for higher dimensions. We note that since M_d is finite dimensional, the Minkowski–Caratheodory convexity theorem [35] implies that a completely positive linear map $\Phi: M_d \rightarrow M_d$ is mixed unitary if and only if Φ is in the convex hull of the unitary transformation maps $X \to UXU^*$.

We shall establish a generalization of the following theorem of Watrous [36] to the setting of operator algebras.

Theorem 2. Let $\Phi: M_d \rightarrow M_d$ be a unital quantum channel. Then for $0 \unicode{x2A7D} p \unicode{x2A7D} 1/ (d^2-1)$, the convex combination of maps given by

$$\begin{align} p \, \Phi\left(X\right) + \left(1-p\right) \frac{\mathrm{Tr}\left(X\right)}{d} I_d \end{align} \tag{ 3 }$$

is a mixed unitary channel.

The completely depolarizing channel $\delta_d : M_d \rightarrow M_d$ is defined as: $\delta_d(X) = d^{-1}\mathrm{Tr} (X) I_d$. As it is also a mixed unitary channel (implemented by any set of (uniformly scaled) unitary operators that form an orthogonal basis in the trace inner product on M_d ), and the set of mixed unitary channels is convex, the theorem is proved by explicitly proving the case $p = 1/(d^2-1)$. That is, theorem 2 is equivalent to the statement that the convex combination $p \Phi + (1-p)\delta_d$ is mixed unitary for $p = \frac{1}{d^2-1}$.

Some initial investigation shows that a naive generalization of theorem 2 does not hold. Consider the following example, which illustrates this point.

Example 3. We have the identity map $\mathrm{id}: M_3\rightarrow M_3$, $\mathrm{id}(X) = X$, and the Werner–Holevo channel [38] on M₃ given by,

$$\begin{align*} W_3^{-}\left(X\right) = \frac{1}{2} \left[\mathrm{Tr}\left(X\right) I_3 - X^t\right], \end{align*}$$

where X^t is the transpose of X. Now consider the channel $\Phi_p : M_3 \rightarrow M_3$ for $0\lt p\unicode{x2A7D} 1$ given by,

$$\begin{align*} \Psi_p\left(X\right) = p X+ \left(1-p\right) W_3^{-}\left(X\right). \end{align*}$$

We claim that this channel is not mixed-unitary for any p. Indeed, first note that any operator-sum representation of $\Psi_p$ will have Kraus operators of the form $K = \alpha I+ A$, where A is an anti-symmetric matrix, I is the identity matrix and α is a constant. (To see this, observe that $\Psi_p$ has a representation of this form, and then note this implies any representation has this form.) As A is a $3\times 3$ anti-symmetric matrix, the eigenvalues are $\lambda, -\lambda, 0$. So the eigenvalues of K are $\alpha+\lambda, \alpha-\lambda, \alpha$. Now if K is a multiple of a unitary, then these eigenvalues must lie on a circle. However, these three numbers are co-linear. Hence λ = 0 and so A = 0. This is true for all Kraus operators and thus it follows that p = 1.

One might expect a naive generalization of the original Watrous Theorem to find that $t \mathrm{id} + (1-t)\Phi$ is mixed unitary for some t, simply replacing δ₃ with the identity map $\mathrm{id}$. But in fact, $W_3^{-}$ is a channel for which $t \mathrm{id} + (1-t) W_3^{-}$ is not mixed unitary for any t < 1. So, simply replacing the depolarizing channel by another unital channel immediately yields that there are channels for which no non-trivial convex combination is mixed unitary.

After some more thought, we were led to view theorem 2 as a special case of a more general phenomena in the context of operator algebras. In particular, in seeking to generalize the theorem, we make the following observations:

• δ_d is the (unique) trace preserving conditional expectation onto the trivial scalar algebra $\mathcal{A} = \mathbb{C} I_d$.
• Every unital channel Φ contains the trivial algebra in its fixed point algebra; $\mathrm{Fix}(\mathcal{A}) \supseteq \mathbb{C} I_d$.
• The unitary group $\mathcal{U}(d)$ inside M_d is the group of unitaries contained in the commutant of the trivial algebra; $(\mathbb{C} I_d)^{\prime} = M_d$.

Following further investigation, we replace the trivial algebra $\mathbb{C} I_d$ with an arbitrary unital operator algebra $\mathcal{A}$, and then we formulated a conjecture on the generalization, which we state and prove as the following result.

Theorem 4. Let $\mathcal{A}$ be any unital operator algebra inside M_d . Let $\mathcal{E}_{\mathcal{A}}$ be the trace preserving conditional expectation onto $\mathcal{A}$, and let $\mathcal{U}_{\mathcal{A}^{^{\prime}}}$ be the group of unitaries contained in the commutant of $\mathcal{A}$. Then for any unital channel Φ whose fixed point algebra contains $\mathcal{A}$, there exists a $p \in (0,1)$ depending only on the algebra $\mathcal{A}$ such that the convex combination

$$\begin{align} p\Phi + \left(1-p\right)\mathcal{E}_{\mathcal{A}}\end{align} \tag{ 4 }$$

is in the convex hull of channels of the form $\Phi_U (X) = UXU^*$ where $U\in \mathcal{U}_{\mathcal{A}^{^{\prime}}}$.

Returning to the example above, the point is, in order to generalize properly, one must restrict the set of channels, Φ, to only those that fix the algebra onto which the conditional expectation projects. In the example, $\mathrm{id}$ is the conditional expectation onto the full matrix algebra, M₃, and in fact there are no non-trivial channels that fix this algebra. It is also the case that the fixed point algebra of $W_3^{-}(X)$ is just the trivial algebra, consisting of scalar multiples of the identity matrix. Thus, there is no unital channel other than δ₃ for which we should expect a Watrous-type theorem to hold for $W_3^{-}$.

We first show how Watrous' theorem 2 yields an asymptotic result for the case of primitive unital channels. We begin with a result that we expect is well-known (see [6] for instance), but for completeness we provide a short proof that makes use of convexity theory (see [35]).

Lemma 12. Let $\mathcal{A}$ be any unital $*$-subalgebra of M_d . Let $\mathcal{E}_{\mathcal{A}}$ be the trace preserving conditional expectation onto $\mathcal{A}$. Then $\mathcal{E}_{\mathcal{A}}$ is a mixed unitary channel.

Proof. Let $\mathcal{U}(\mathcal{A}^{^{\prime}})$ be the unitary group of the commutant algebra $\mathcal{A}^{^{\prime}} = \{X\in M_d: A X = X A, \forall A\in \mathcal{A}\}.$ It follows that the conditional expectation can be written as follows for all $X\in M_d$:

$$\begin{align*} \mathcal{E}_{\mathcal{A}}\left(X\right) = \int_{U \in \mathcal{U}\left(\mathcal{A}^{^{\prime}}\right)} UXU^* \mathrm{d}\mu\left(U\right) ,\end{align*}$$

where $\mu(U)$ is the normalized Haar measure on $\mathcal{U}(\mathcal{A}^{^{\prime}})$. Indeed, it is easy to see that this integral operator is trace preserving, projects onto $\mathcal{A}$, and satisfies the other conditional expectation properties from the invariance of the Haar measure. So by uniqueness the map is $\mathcal{E}_{\mathcal{A}}$.

Now as the commutant $\mathcal{A}^{^{\prime}}$ is a finite dimensional subalgebra, the group $\mathcal{U}(\mathcal{A}^{^{\prime}})$ is closed (and hence compact) and so the convex hull of the set $\{UXU^*: U \in \mathcal{U}(\mathcal{A}^{^{\prime}}) \}$ is a compact (and hence closed) convex set by the Minkowski-Caratheodory convexity theorem. Thus $\mathcal{E}_{\mathcal{A}}(X)$ lies in the convex hull of mappings of the form $UXU^*$, with $U \in \mathcal{U}(\mathcal{A}^{^{\prime}})$, and so $\mathcal{E}_{\mathcal{A}}$ is a mixed unitary map. □

Remark 13. Note that from the above result, it is clear that the completely depolarizing channel $\delta_d(X) = d^{-1}\mathrm{Tr} (X) I_d$, which is the trace preserving conditional expectation onto the trivial algebra $\mathcal{A} = \{\mathbb{C} I\}$, is a mixed unitary map. A concrete representation of this map can be written down:

$$\begin{align*}\delta_d\left(X\right) = d^{-1}\mathrm{Tr} \left(X\right) I_d = \frac{1}{d^2} \sum_{a,b = 0}^{d-1} W_{a,b} X W_{a, b}^{\,*}, \end{align*}$$

where $W_{a,b}$ are the Weyl-Heisenberg unitaries defined by

$$\begin{align*}W_{a,b} = S^a D^b, 0\unicode{x2A7D} a, b\unicode{x2A7D} d-1, \end{align*}$$

and $S = \sum_{j = 1}^d E_{j+1(\mathrm{mod}\, d), j}\in M_d$ is the forward cyclic shift operator and $D = \sum_{j = 1}^d \omega^j E_{j,j}\in M_d$ is the 'clock operator' with $\omega = \text{exp}(2\pi i/d)$.

We use the above result to prove the following. Let us denote MU(d) to be the set of all mixed-unitary channels on M_d , which note is a closed convex set of linear maps. See [29] for basic properties of the completely bounded distance measure.

Lemma 14. Let $\Phi:M_d\rightarrow M_d$ be any unital quantum channel. Then

$$\begin{align*} \displaystyle {{\text{liminf}} _{n\to\infty} } \, d_{CB}\left( \Phi^n, \text{MU(d)}\right) = 0,\end{align*}$$

where $d_{CB}( \Phi^n, \text{MU(d)})$ is the completely bounded distance of $\Phi^n$ from the closed convex set $\text{MU(d)}$.

Proof. By Kuperberg's Theorem (see theorem 2.1 in [22] ), for a unital completely positive map Φ, there is a subsequence $n_1, n_2, \cdots$ such that

$$\begin{align*}\lim_{j\to\infty} \Phi^{n_j} = \mathcal{E}_\Phi,\end{align*}$$

where $\mathcal{E}_\Phi$ is a unique idempotent map. Note that for trace-preserving completely positive maps, such a map also exists (see proposition 6.3 in [39]).

Now note that since Φ is trace preserving, $\mathcal{E}_\Phi$ is also trace preserving. And hence its range $\textrm{Range} ( \mathcal{E}_\Phi)$ is a C$^*$-algebra. Indeed, let $a \in \textrm{Range} (\mathcal{E}_\Phi)$, then from Schwarz inequality it follows that

$$\begin{align*}\mathcal{E}_\Phi \left(aa^*\right)\unicode{x2A7E} \mathcal{E}_\Phi\left(a\right)\mathcal{E}_\Phi\left(a^*\right) = aa^*.\end{align*}$$

And using the trace preservation property and faithfulness of trace, we get that $\mathcal{E}_\Phi(aa^*) = aa^*$. Thus, $aa^*\in \textrm{Range} ( \mathcal{E}_\Phi)$. This means $\mathcal{E}_\Phi (aa^*) = \mathcal{E}_\Phi (a)\mathcal{E}_\Phi (a^*) = aa^*$. And hence a is in the multiplicative domain of $\mathcal{E}_\Phi$ ([7]). From here, it follows that if $a, b\in \textrm{Range} (\mathcal{E}_\Phi)$, then $\mathcal{E}_\Phi(ab) = ab$. Hence, $ab\in \textrm{Range} (\mathcal{E}_\Phi)$, showing $\textrm{Range} (\mathcal{E}_\Phi)$ is a C$^*$-algebra.

Now the result follows from lemma 12. □

For a special class of channels (e.g. see [1, 33, 34]), one can make a stronger statement.

Definition 15. A unital channel $\Phi:M_d\rightarrow M_d$ is primitive if it is irreducible (i.e. $\Phi(P)\unicode{x2A7D} \lambda P$ for some projection P implies P = 0 or P = I) and it has a trivial peripheral spectrum (i.e. $\text{spec}(\Phi)\cap \mathbb{T} = \{1 \}$, where $\mathbb{T}$ is the unit circle in the complex plane).

Theorem 16. For every primitive unital channel $\Phi:M_d\rightarrow M_d$, there is a finite $k\in \mathbb{N}$ such that $\Phi^k$ is mixed unitary, and subsequenctly for every $l\unicode{x2A7E} k$, $\Phi^l$ is mixed unitary.

Proof. It follows from the proof of the previous result that for a primitive unital channel Φ, the conditional expectation $\mathcal{E}_\Phi$ described above is the completely depolarizing channel δ_d . Now by Watrous's theorem (2) there is a ball around δ_d where every unital channel is mixed-unitary. So from the subsequence $n_1, n_2, \cdots,$ if we take sufficiently large n_i 's, the maps $\Phi^{n_i}$ must fall in the ball around δ_d . Hence there is a $k\in \mathbb{N}$ such that $\Phi^k$ is in this ball and it is mixed unitary.

The second statement follows easily from the above argument and the CB norm estimate:

$$\begin{align*}||\Phi^{k+r}-\delta_d ||_{CB} = ||\Phi^r \left(\Phi^k-\delta_d\right)||_{CB}\unicode{x2A7D} || \left(\Phi^k-\delta_d\right)||_{CB} , \end{align*}$$

which also uses the fact that $\Phi \circ \delta_d = \delta_d$ as Φ is unital. □

In what follows, we will show how theorem 4 allows us to prove an analogous result for all unital channels.

Next we derive some properties of peripheral eigenvalues for irreducible unital channels.

Let us first observe that a unital channel Φ is irreducible if and only if its fixed point algebra $\mathrm{Fix}(\Phi)$ is just the scalar algebra, $\mathcal{A} = \mathbb{C} I$. Indeed, if Φ is irreducible, then only the trivial projections are fixed by Φ, and hence its fixed point algebra (which is spanned by its projections as a von Neumann algebra) must be trivial. Conversely, if the fixed point algebra for Φ is trivial and P is a projection with $\Phi(P) \unicode{x2A7D} \lambda P$, then $\Phi(P)$ is supported on the range of P, which for a unital channel implies (as proved in [20]) that in fact $\Phi(P) = P$ if it is non-zero, and hence P = 0 or P = I.

In the following result, we denote the set of unital channels that fix a given algebra $\mathcal{A}$ by $\mathcal{F}(\mathcal{A})$. Evidently this set has the structure of a convex semigroup under composition of maps. It is also $\ast$-closed, in the sense that a map is in the set if and only if its dual map is as well (which can be seen as a consequence of the fixed point theorem for unital channels [20]).

Lemma 17. Let $\mathcal{A}$ be a unital subalgebra of M_d that is unitarily equivalent to $\oplus_{k = 1}^r I_{m_k} \otimes M_{n_k}$, and let $\mathcal{F}(\mathcal{A})$ be the semigroup of unital channels on M_d that fix $\mathcal{A}$. Let $\widehat{\mathcal{A}} = \oplus_{k = 1}^r \mathbb{C} I_{m_k}$, with associated semigroup $\mathcal{F}(\widehat{\mathcal{A}})$ of unital channels on $M_{(\sum_k m_k)}$ that fix $\widehat{\mathcal{A}}$.

Then, there is a convex $\ast$-semigroup isomorphism $\alpha : \mathcal{F}(\mathcal{A}) \rightarrow \mathcal{F}(\widehat{\mathcal{A}})$ with the property that $\alpha(\Phi) \in \mathcal{F}(\widehat{\mathcal{A}})$ is mixed unitary if and only if $\Phi \in \mathcal{F}(\mathcal{A})$ is mixed unitary.

Proof. Let $\Phi \in \mathcal{F}(\mathcal{A})$ with Kraus operators $\{K_i\}_{i = 1}^n$. Since Φ fixes $\mathcal{A}$, we have $\mathcal{A} \subseteq \mathrm{Fix}(\Phi) = \{K_i \}^{\prime}$, and so $K_i \in \mathcal{A}^{^{\prime}}$. Hence there exists a unitary $U\in M_d$ and matrices $K_{ik}\in M_{m_k}$ such that $U^*K_i U = \oplus_{k = 1}^r K_{ik}\otimes I_{n_k}$ for all i.

Then $\alpha(\Phi)$ is defined to be the map whose Kraus operators are $\widehat{K_i}: = \oplus_{k = 1}^r K_{ik}$, which, as a notational convenience, we sometimes write as $\widehat{K_i} = \alpha(K_i)$. To show that $\alpha(\Phi)$ fixes $\widehat{\mathcal{A}}$, we note that the Kraus operators of $\alpha(\Phi)$ always lies in the algebra $\oplus_{k = 1}^r M_{m_k}$, and that $\widehat{\mathcal{A}}$ thus necessarily commutes with $\widehat{K_i}$; hence it is contained in the fixed point algebra.

It is easy to see that the image of α does not depend on a particular operator-sum representation of Φ, and, moreover, that an operator-sum representation of Φ is minimal in terms of number of Kraus operators if and only if the image of the Kraus operators under α is a minimal representation of $\alpha(\Phi)$.

It is also clear that α is a $*$-homomorphism, since for any Φ, $\Psi \in \mathcal{A}$ with respective Kraus operators $U^*K_i U = \oplus_{k = 1}^r K_{ik}\otimes I_{n_k}$ and $U^*L_i U = \oplus_{k = 1}^r L_{ik}\otimes I_{n_k}$, we have that $\Phi\circ \Psi$ has Kraus operators $U\bigl(\oplus_{k = 1}^r K_{ik}L_{i^{^{\prime}}k}\otimes I_{n_k}\bigr)U^*$. The image under α of these operators are $\oplus_{k = 1}^r K_{ik}L_{i^{^{\prime}}k}$, which are exactly the Kraus operators of $\alpha(\Phi)\circ \alpha(\Psi)$.

To see that α is an isomorphism, let $\Phi,\Psi \in \mathcal{F}(\mathcal{A})$, and suppose Φ has a minimal set of Kraus operators given by $\{K_i = U\bigl( \oplus_{k = 1}^r K_{ik}\otimes I_{n_k}\bigr)U^*\}_{i = 1}^n$ and Ψ has a minimal set of Kraus operators $\{L_i = U\bigl(\oplus_{k = 1}^r L_{ik}\otimes I_{n_k}\bigr)U^*\}_{i = 1}^{n^{^{\prime}}}.$ Then $\alpha(\Phi)$ has Kraus operators $\{\widehat{K_i} = \oplus_{k = 1}^r K_{ik}\}_{i = 1}^n$ and $\alpha(\Psi)$ has $\{\widehat{L_i} = \oplus_{k = 1}^r L_{ik}\}_{i = 1}^{n^{^{\prime}}}$. If $\alpha(\Phi) = \alpha(\Psi)$, then we have $\{\widehat{K_i}\}_{i = 1}^n$ and $\{\widehat{L_i}\}_{i = 1}^{n^{^{\prime}}}$ are two different minimal Kraus representations of the same channel, so $n = n^{^{\prime}}$. Hence there exists a scalar unitary matrix $V = (v_{ij})$ such that $\widehat{L_i} = \sum_j v_{ij}\widehat{K_j}$ and so $L_{ik} = \sum_j v_{ij}K_{jk}.$ Thus

$$\begin{eqnarray*} L_i = U\left(\oplus_{k = 1}^r L_{ik}\otimes I_{n_k}\right)U^* & = & U\left(\oplus_{k = 1}^r \left(\sum_{j = 1}^n v_{ij}K_{jk}\right)\otimes I_{n_k}\right)U^* \\ & = & U\left(\sum_j v_{ij} \left(\oplus_{k = 1}^r K_{jk}\otimes I_{n_k}\right)\right)U^* = \sum_j v_{ij}K_j , \end{eqnarray*}$$

and so $\{L_i\}$ and $\{K_i\}$ are two different representations of the same channel, giving $\Phi = \Psi$.

Finally, $\alpha(\Phi)$ is a unitary adjunction channel if and only if Φ is; since $\oplus_{k = 1}^r U_k \otimes I_{j_k}$ is unitary if and only if each U_k is unitary, which in turn is equivalent to $\oplus_{k = 1}^r U_k$ being unitary. So, in one direction, if $\Phi = \sum_i p_i \mathrm{ad}_{U_i}$ expresses Φ as a convex combination of unitary adjunction maps (where $\mathrm{ad}_U(X) = UXU^*$), the (convex) linearity of α guarantees that

$$\begin{align*} \alpha\left(\Phi\right) = \sum_i p_i \alpha\left(\mathrm{ad}_{U_i}\right) \end{align*}$$

expresses $\alpha(\Phi)$ as a convex combination of the unitary adjunctions $\alpha(\mathrm{ad}_{U_i})$. In the other direction, suppose

$$\begin{align*} \alpha\left(\Phi\right) = \sum_i p_i \mathrm{ad}_{\widehat{U_i}} \end{align*}$$

for some unitaries $\widehat{U_i}$. Since $\alpha(\Phi)$ fixes the algebra $\widehat{\mathcal{A}} = \oplus_{k = 1}^r I_{m_k}$, it must be that each $\widehat{U_i} \in \widehat{\mathcal{A}}^{^{\prime}} = \oplus_{k = 1}^r M_{m_k}$ and hence $\widehat{U_i} = \oplus_{k = 1}^r U_{ik}$ for some unitaries U_ik on each block. If we define $U_i = U\bigl(\oplus_{k = 1}^r U_{ik}\otimes I_{n_k}\bigr)U^* \in \mathcal{A}^{\prime}$, it is clear that $\alpha(\Phi)$ is now the image of $\Psi: = \sum_i p_i \mathrm{ad}_{U_i}$, which is mixed unitary. Since α is an isomorphism, and $\alpha(\Phi) = \alpha(\Psi)$, it must in fact be that $\Phi = \Psi$ and hence is mixed unitary. □

Remark 18. Notice that if $\mathcal{A}$ is the fixed point algebra of Φ, i.e. the largest unital algebra fixed by Φ, then the algebra generated by its Kraus operators K_i is $\mathcal{A}^{^{\prime}}$. So $\alpha(\Phi)$ has Kraus operators that generate the algebra $\alpha(\mathcal{A}) = \oplus_{k = 1}^r M_{m_k}$, and so the fixed point algebra of $\alpha(\Phi)$ is the abelian algebra $\oplus_{k = 1}^r \mathbb{C} I_{m_k}$. Also notice that the channels $\Phi_k$, with Kraus operators $\{K_{ik}\}$ are irreducible. Thus, without loss of generality, we will prove our result for channels with abelian fixed point algebra, as any unital channel is identified with a channel that has abelian fixed point algebra, and where the identification carries through the relevant properties (i.e. commutes with powers and preserves mixed unitarity).

We next consider the peripheral spectrum for a map $\Phi : M_d \rightarrow M_d$, which is the set

$$\begin{align*} \left\{X\in M_d \,\, | \,\, \Phi\left(X\right) = \lambda X \,\, \mathrm{for\, some} \, |\lambda| = 1\right\}. \end{align*}$$

In the case of an irreducible unital channel, there is a positive integer m such that the peripheral spectrum is $\{\omega^i\}_{i = 0}^{m-1}$ for some primitive m^th root of unity (see for instance theorem 6.6 from [39]). Further, as shown in [32] (theorem 2.5), for a unital channel Φ, the algebra generated by all peripheral eigen-operators for Φ is equal to the algebra $\mathcal{M}_{\Phi^\infty}$, which is defined as the decreasing intersection of the multiplicative domains $\mathcal{M}_{\Phi^k}$ for $\Phi^k$, $k\unicode{x2A7E} 1$; in particular, the peripheral spectrum of $\Phi^k$ is a subset of the peripheral spectrum of Φ.

The following useful fact for us comes as a simple consequence of the spectral mapping theorem, from which it follows that the spectrum of $\Phi^m$ consists of the elements of the spectrum of Φ raised to the mth power.

Lemma 19. Suppose Φ is an irreducible unital channel with peripheral spectrum $\{\omega^i\}_{i = 0}^{m-1}$ for some primitive $m\mathrm{th}$ root of unity. Then $\Phi^m$ has no non-trivial peripheral spectrum; that is, $\mathrm{spec}(\Phi^m)\cap \mathbb{T} = \{1 \}$.

We next recall basic features of peripheral eigenvalues, with a short proof for completeness.

Lemma 20. Let $\Phi : M_d \rightarrow M_d$ be a unital channel, and let X be a peripheral eigenvector for Φ: $\Phi(X) = \lambda X$ for some $|\lambda| = 1$. Then $K_i X = \lambda XK_i$ for all Kraus operators K_i , and so if X is a peripheral eigenvector for Φ with eigenvalue λ, then we have

$$\begin{align*} \Phi\left(XA\right) = \lambda X \Phi\left(A\right) \quad \quad \Phi\left(AX\right) = \overline{\lambda}\Phi\left(A\right)X , \end{align*}$$

for all $A\in M_d$.

Proof. Define $A_i = K_i X - \lambda XK_i$. Then we have,

$$\begin{align*}\sum_i A_i A_i^* & = \sum_i K_i X X^* K_i^* -\overline{\lambda}\sum_i K_i XK_i^*X^* - \lambda X \sum_i K_i X^*K_i^* + |\lambda|^2 X \sum_i K_iK_i^* X^* \\ & = \Phi\left(XX^*\right) - XX^*, \end{align*}$$

where we use the fact that $\Phi(X) = \lambda X$, $\Phi(X^*) = \overline{\lambda} X^*$, and that Φ is unital. Then, by trace preservation, we have that

$$\begin{align*} \sum_i \mathrm{Tr}\left(A_iA_i^*\right) = \mathrm{Tr}\left(\Phi\left(XX^*\right)\right) - \mathrm{Tr}\left(XX^*\right) = 0 , \end{align*}$$

and hence each $A_i = 0$. The final statement immediately follows. □

We also need the following characterization of peripheral eigenvectors in the commutative fixed point algebra case. We note a possible connection between this result and results obtained recently in [2].

Lemma 21. Let Φ be a unital channel with fixed point algebra unitarily equivalent to $\oplus_{k = 1}^r \mathbb{C} I_{m_k}$. Let X be a peripheral eigenvector. Then one of the two following cases holds:

• 1.  
  $X = \oplus_{k = 1}^r X_k$ where each X_k is a peripheral eigenvector for the irreducible channel $\Phi_k$ obtained by restricting the Kraus operators of Φ to the k^th diagonal block.
• 2.  
  There exists $j,k$ such that $m_j = m_k$, and there is a unitary U on $\mathbb{C}^{m_j}$ such that $\Phi_j = \mathrm{ad}_U \circ \Phi_k \circ \mathrm{ad}_{U^*}$.

Proof. Up to unitary equivalence, the fixed point algebra has minimal central (orthogonal) projections $P_i = \oplus_{k = 1}^r \delta_{ik} I_{i_k}$ with $\sum_i P_i = I$. As these are fixed points of Φ, we have by lemma 20 that $\Phi(P_kXP_j) = P_k \Phi(X) P_j$ for all X and $k,j$. In particular, applying this to the peripheral eigenvector X with eigenvalue λ, we get

$$\begin{align*} \Phi\left(P_kX P_j\right) = \lambda P_kX P_j \end{align*}$$

for all pairs $j,k$. That is, $P_kX P_j$ is also a peripheral eigenvector for Φ with eigenvalue λ.

Now, lemma 20 also shows that, for any peripheral eigenvector X we have

$$\begin{align*} \Phi\left(XX^*\right) = |\lambda|^2 XX^* = XX^*, \end{align*}$$

and so $XX^*$ must be in the span of the P_i . Hence the same is true for $X^*X$, $P_k X P_j X^* P_k$ and $P_k X^* P_j X P_k$.

Thus we can find scalars c_i such that

$$\begin{align*} P_kXP_jX^*P_k = \sum_i c_i P_i , \end{align*}$$

which yields after multiplying on the left or right by P_k that

$$\begin{align*} P_k XP_jX^*P_k = c_k P_k. \end{align*}$$

If we let X_kj be the operator corresponding to the (k, j) block in the decomposition determined by the $\{P_j\}$, which is $P_k X$ restricted to the range of P_j , then we have that

$$\begin{align} X_{kj}X^*_{kj} = c_k I_{m_k} \quad \text{normal}\;\mathrm{and} \quad X_{kj}^*X_{kj} = c_j I_{m_j}. \end{align} \tag{ 12 }$$

There are two possibilites: either $c_k = c_j = 0$, or both scalars are non-zero and X_kj is a (non-zero) multiple of a unitary (and $m_j = m_k$). Thus, in this block matrix form, any peripheral eigenvector has the form,

$$\begin{align*} X = \sum_{i_1,i_2} E_{i_1 i_2} \otimes X_{i_1 i_2}, \end{align*}$$

where each $X_{i_1 i_2}$ is either 0 or a (non-zero) multiple of a unitary. Moreover, we know by lemma 20 that $K_i X = \lambda X K_i$, and so we have that

$$\begin{align*} K_{ij}X_{jk} = \lambda X_{jk} K_{ik} \end{align*}$$

for all i and all (j, k). In the case that X_jk is non-zero, we therefore have, after multiplying both sides on the right by $X_{jk}^*$ and using equation (12) with the roles of k and j reversed,

$$\begin{align*} K_{ij} = \frac{1}{c_j}\lambda X_{jk}K_{ik}X_{jk}^*. \end{align*}$$

Since $\frac{X_{jk}}{\sqrt{c_j}}$ is unitary, and $|\lambda| = 1$, this expresses K_ij as a unitary conjugation of K_ik for all i; that is, if $\Phi_i$ is the channel whose Kraus operators are $\{K_{ij}\}_{j = 1}^n$, then $\Phi_j = \mathrm{ad}_U \circ \Phi_k \circ \mathrm{ad}_{U^*}$ with $U = \sqrt{\frac{\lambda}{c_j}}X_{jk}$. □

Combining these last results with Kuperberg's Theorem [22] and our main result from the last section, allows us to prove the following.

Theorem 22. Let Φ be a unital channel. Then there exists an integer k > 0 such that $\Phi^k$ is mixed unitary.

Proof. By lemma 17, perhaps by replacing Φ with $\alpha(\Phi)$, we can without loss of generality assume Φ has a commutative fixed point algebra. Then, lemmas 19 and 21 show that a high enough power, $M\unicode{x2A7E} 1$ say, of Φ has no non-trivial peripheral spectrum; for instance, M can be taken as the lowest common multiple of the m's from lemma 19 applied to a representative from each of the irreducible channel unitary equivalence classes found in lemma 21.

Thus, $\Phi^M$ is a unital channel with no non-trivial peripheral spectrum, and so its peripheral algebra is just its fixed point algebra. We can now apply Kuperberg's Theorem in this case to find a subsequence $\{k_i\}$ such that $(\Phi^M)^{k_i} = \Phi^{M k_i} \rightarrow \mathcal{E}_{\mathcal{A}}$ where $\mathcal{A}$ is the fixed point algebra of $\Phi^M$. By theorem 4, there is a ball around $\mathcal{E}_{\mathcal{A}}$ whose intersection with $\mathcal{F}(\mathcal{A})$, the set of unital channels with fixed point algebra $\mathcal{A}$, consists only of mixed unitaries, and so any channel in $\mathcal{F}(\mathcal{A})$ sufficiently close to $\mathcal{E}_{\mathcal{A}}$ is mixed unitary. Therefore, it follows that there is a k_N such that, for all i > N, the channel $(\Phi^M)^{k_j}$ is sufficiently close to $\mathcal{E}_{\mathcal{A}}$ that it is mixed unitary, and this completes the proof. □

Remark 23. Notice that in order to obtain this result, we cannot use Kuperberg's Theorem directly with the conditional expectation onto the peripheral algebra; this is because the ball of mixed unitaries we obtain around $\mathcal{E}_{\mathcal{A}}$ is in the relative interior of $\mathcal{F}(\mathcal{A})$, the set of all unital channels with fixed point algebra $\mathcal{A}$. So if $\Phi^M$ only has peripheral algebra $\mathcal{A}$, but not fixed point algebra $\mathcal{A}$, although $\Phi^{M k_i} \rightarrow \mathcal{E}_{\mathcal{A}}$, it may approach from outside the relative interior $\mathcal{F}(\mathcal{A})$ where the Theorem does not apply.

Remark 24. We also draw the attention of the reader to a conjecture called the 'Asymptotic Quantum Birkhoff Conjecture', which asks whether for a unital quantum channel $\Phi: M_n \rightarrow M_n$, it holds that,

$$\begin{align*} \lim_{k \rightarrow \infty} d_{CB} \left(\Phi^{\otimes k}, MU\left(M_n^{\otimes k}\right)\right) = 0, \end{align*}$$

where, as above, d_CB is the completely bounded distance of $\Phi^{\otimes k}$ to the set of mixed unitary maps on $M_n^{\otimes k}$. The conjecture was resolved in the negative by Haagerup and Musat ([13]). They introduced a new class of maps called factorizable maps and showed that maps which are not factorizable, fail to satisfy the above conjecture. In essence, this means that not every unital channel, after taking tensor powers with itself, becomes mixed unitary even if we take larger and larger tensor powers. In contrast, lemma 14 shows that every unital channel 'asymptotically becomes' mixed unitary. Quite significantly, theorem 22 goes further and uncovers an interesting aspect of unital channels the contrasts with tensor powers: it says under composition, every unital channel becomes mixed unitary after finitely many applications.

Let G be an Abelian group. We now fix some notation. Since G is Abelian we use additive notation and hence 0 is the group identity. The group G^d is the Cartesian product of d copies of G. Reminiscent of the standard basis in linear algebra, if $G = \mathbb{Z}_m$ or $G = \mathbb{Z}$, then e_k denotes the element in G^d consisting of an d-tuple of elements of G where the kth element is 1 and all other elements are 0. So for instance, e_i  – e_j is the d-tuple of elements of G where the ith element is 1, the jth element is −1, and all of other n − 2 elements are 0.

We let $\widehat{G}$ be the set of all group homomorphisms from G to $\mathbb{T}$, the unit circle in the complex plane. The set $\widehat{G}$ is a group under multiplication and is called the dual group. The Abelian groups $\mathbb{Z}^d$ and $\mathbb{T}^d$ are duals to one another and any finite Abelian group is self-dual. While these are well known results in abstract harmonic analysis, it is illustrative to prove one of these results. Suppose h is a homeomorphism from $\mathbb{Z}^d$ to $\mathbb{T}$, then h is completely determined by the d values $h(e_1)$, $h(e_2)$,..., $h(e_n)$. The map $h\to (h(e_1), h(e_2),..., h(e_n))$ is a group isomorphism between $\widehat{\mathbb{Z}^d}$ and $\mathbb{T}^d$ proving that $\widehat{\mathbb{Z}^d} = \mathbb{T}^d$.

Let µ be any measure on an Abelian group G, then the Fourier transform of µ is the complex valued function on $\widehat{G}$ defined as follows: $\widehat{\mu}(\chi) = \int_G \chi(g) \mathrm{d}\mu (g)$. A complex-valued function on $\widehat{G}$ is said to be positive definite if it is the Fourier transform of a measure on G.

We can characterize the convex hulls of rank one correlation matrices in both the real and the complex cases in terms of positive definite functions. The real version of the result which we present first is essentially equivalent to [4, proposition 2.1] and [28, theorem 7].

Theorem 30. Let C be an d × d real matrix. Then C is in the convex hull of the real rank one correlation matrices if and only if there exists a positive definite function $f:\mathbb{Z}_2^d\to \mathbb{R}$ with the following properties:

• 1.  
  $f(0) = 1$
• 2.  
  $f(e_i-e_j) = c_{ij}$ for $1\unicode{x2A7D} i\lt j \unicode{x2A7D} d$

The complex version of this theorem, which we now state, appears to be new.

Theorem 31. Let C be an d × d complex matrix. Then C is in the convex hull of the complex rank one correlation matrices if and only if there exists a positive definite function $f:\mathbb{Z}^d\to \mathbb{C}$ with the following properties:

• 1.  
  $f(0) = 1$
• 2.  
  $f(e_i-e_j) = c_{ij}$ for $1\unicode{x2A7D} i\lt j \unicode{x2A7D} d$.

We can combine these two versions into a common generalization as follows.

Theorem 32. Let C be an d × d complex matrix. Let G be any topologically closed subgroup of $\mathbb{T}$. Then C is in the convex hull of the rank one correlation matrices with all entries in G if and only if there exists a positive definite function $f:\widehat{G}^d\to \mathbb{C}$ with the following properties:

• 1.  
  $f(0) = 1$
• 2.  
  $f(e_i-e_j) = c_{ij}$ for $1\unicode{x2A7D} i\lt j \unicode{x2A7D} d$

Setting $G = \mathbb{Z}_2$ in theorem 32 gives us theorem 30 and setting $G = \mathbb{T}$ in theorem 32 gives us theorem 31. Hence we only need prove theorem 32.

Proof. Let G be any topologically closed subgroup of $\mathbb{T}$. If v is any n-vector all of whose entries are in G, then let δ_v denote the probability measure on G^d satisfying $\delta_v(\{v\}) = 1$. Let $\widehat{\delta}_v$ be the corresponding positive definite function (i.e. for any $\chi \in \widehat{G}^d$, $\widehat{\delta}_v(\chi) = \int_G \chi(g) d\delta_v = \chi(v))$. If $\chi = (c_1,c_2,{\ldots},c_d)\in \widehat{G}^d$ and $v = (v_1,v_2,{\ldots},v_d)\in G^d$, then $\chi(v) = \prod_{k = 1}^dv_k^{c_k}$. Hence $\widehat{\delta}_v(e_i-e_j) = v_iv_j^{-1} = v_i\overline{v_j}$ since $|v_j| = 1$. Therefore for all $i,j$, $\widehat{\delta}_v(e_i-e_j)$ is the (i, j)th entry of the matrix $vv^*$. Now if C is in the convex hull of the rank one correlation matrices with all entries in G, there exists $\{\lambda_i\}_i$ positive numbers summing to one and $\{v_i\}_i$ n-vectors having all elements in G such that $C = \sum_i\lambda_iv_iv_i^*$. It follows from linearity that the Fourier transform of the probability measure $\sum_i \lambda_i\delta_{v_i}$ is the f which satisfies all the hypotheses of the theorem. The converse follows by reversing the steps of this argument. For the $G = \mathbb{T}$ case, we note that the set of all probability measures on $\mathbb{T}$ is weak-$*$ compact by the Banach–Alaoglu theorem and hence is the closed convex hull of the point measures on $\mathbb{T}$ by the Krein–Milman theorem. The result now follows using a similar argument to the topologically closed subgroup case. □

We can use theorem 31 to construct an improvement on theorem 28. We begin with the following lemma which gives a useful example of a positive definite function on the integers.

Lemma 33. Let c be a complex number of modulus less than or equal to one. Then the function $f_c(n):\mathbb{Z}\to \mathbb{C}$ defined as

$$\begin{align*}f_c\left(n\right) = \begin{cases} c^n & n\unicode{x2A7E} 0 \\ \overline{c}^{\vert n \vert}\ & n<0 \end{cases} \end{align*}$$

is positive definite.

Proof. Note that the Mobius transformation $g(z) = \frac{1}{1-z}$ maps the closed unit disk of the complex plane to the half plane $\{z: Re(z)\unicode{x2A7E} \frac{1}{2}\}$. It follows from this that when $\vert c\vert\unicode{x2A7D} 1$ and c ≠ 1, then $\widehat{f_c}(e^{i\theta}) = \sum_{k\in \mathbb{Z}}f_c(k)e^{ik\theta} = -1+\frac{1}{1-ce^{i\theta}}+ \frac{1}{1-ce^{-i\theta}}\unicode{x2A7E} 0$. Hence f_c is positive definite. The function f₁ is the Fourier transform of the point measure at zero and hence is positive definite. □

This has some important implications for correlation matrices.

Corollary 34. Let $v = (v_1,v_2,..,v_d)\in \mathbb{C}^d$ with $\Vert v\Vert_{\infty}\unicode{x2A7D} 1$ and let M(v) denote the d × d matrix having the same off-diagonal entries as $vv^*$ and all diagonal entries equal to one. Then M(v) is in the convex hull of the complex rank one correlation matrices.

Proof. It follows from the previous lemma that $f_{v_k}$ is a positive definite function on $\mathbb{Z}$. Therefore a simple product measure argument shows us that $f(n_1,n_2,{\ldots},n_d) = \prod_{k = 1}^d f_{v_k}(n_k)$ is a positive definite function $f:\mathbb{Z}^d\to \mathbb{C}$. Then $f(0) = 1$ and $f(e_i-e_j) = f_{v_i}(1)f_{v_j}(-1) = v_i\overline{v_j}$. Our result now follows from theorem 31. □

Corollary 35. Let C be a rank r complex correlation matrix. Then $\frac{1}{r}C+\frac{r-1}{r}I$ is in the convex hull of complex rank one correlation matrices.

Proof. We must have vectors $\{v_k\}_{k = 1}^{r}$ such that $C = \sum_{k = 1}^r v_kv_k^*$; then all these vectors must satisfy $\Vert v_k\Vert_{\infty}\unicode{x2A7D} 1$. Since $\frac{1}{r}C+\frac{r-1}{r}I = \frac{1}{r}\sum_{k = 1}^r M(v_k)$, our result now follows from corollary 34. □

Remark 36. We note that this can be viewed as the complex version of [28, theorem 7] which gave an identical result for real correlation matrices. It was observed in [28] that [28, theorem 7] is not optimal at least in small dimensions, and it is likely that the same is true for corollary 35. We note that any extreme point of the set of d × d correlation matrices has rank at most $\lfloor \sqrt{d} \rfloor$ and hence for any d × d complex correlation matrix C, we have that $\frac{1}{\lfloor \sqrt{d} \rfloor}C+\frac{\lfloor \sqrt{d} \rfloor-1}{\lfloor \sqrt{d} \rfloor}I$ is in the convex hull of complex rank one correlation matrices. This result is an improvement on [10, proposition 4.1].