Is the composition of two extremal channels also extremal?

Question

In this question, I follow the terminology and notation of the book of Watrous, most notably chapter two.

Extremal channels

An extremal channel $\Phi(X) \in C(\mathcal{X},\mathcal{Y})$ is a channel that can not be written as a convex combination of other channels. In other words, $\Phi$ is an extremal channel if and only if the equality \begin{equation} \Phi(X) = \lambda \Phi_{1}(X) + (1-\lambda)\Phi_{2}(X), \end{equation} for $\lambda \in (0,1)$ implies that $\Phi_{1} = \Phi_{2} = \Phi$.

Question

Given two extremal channels $\Lambda_{1} \in C(\mathcal{X},\mathcal{Y})$ and $\Lambda_{2} \in C(\mathcal{Y},\mathcal{Z})$, is the composition $\Lambda_{12} \in C(\mathcal{X},\mathcal{Z})$ defined by

$$ \Lambda_{12} = \Lambda_{2} \circ \Lambda_{1}, $$ also an extremal channel? Or are there extra criteria that the maps should adhere to for the composition to be an extremal channel, too?

In other words, suppose that $\Lambda_{12}(X) = \lambda\Lambda_{12}^{1}(X) + (1-\lambda)\Lambda_{12}^{2}(X)$, does that imply that $\Lambda_{12} = \Lambda_{12}^{1} = \Lambda_{12}^{2}$?.

Do $\Lambda_{12}^{1}$ and/or $\Lambda_{12}^{2}$ even necessarily need to have the structure of mapping $L(\mathcal{X})\rightarrow L(\mathcal{Y}) \rightarrow L(\mathcal{Z})$? (Insofar that this is not an ill-posed question).

Some thoughts

As is shown in the aforementiond book (Thm. $2.31$, page $96$), a channel represented by linearily independent Kraus operators $\{A_{k}\}_{k}$ is an extremal channel if and only if the collection $\{A_{k}^{\dagger}A_{k'}\}_{kk'}$ is also linearily independent.

Let $\{A_{k}: A_{k} \in L(\mathcal{X},\mathcal{Y})\}_{k}$ and $\{B_{l}\in L(\mathcal{Y},\mathcal{Z})\}_{l}$ be such Kraus operators for $\Phi_{1}$ and $\Phi_{2}$, respectively.

The collection $\{B_{l}A_{k}: B_{l}A_{k} \in L(\mathcal{X},\mathcal{Z})\}_{kl}$ forms a valid set of Kraus operators representing $\Phi$. However, these might not be linearily independent, which means that we cannot invoke the above criterium for extremal channels. One could take only a spanning but linearily independent subset and invoke the criterium, but I am not sure if this is a properly justified step. Even more so, such a set $\{A^{\dagger}_{k'}B^{\dagger}_{l'}B_{l}A_{k}\}_{(kl),(k'l')}$ is not necessarily linearily independent, I believe.

Answer 1

No, composition of two extremal channels may fail to be extremal.

Below, we construct a counterexample from an amplitude damping channel and an amplitude raising channel. The idea is to first apply the amplitude damping channel $\mathcal{E}$ which maps the entire Bloch sphere to $|0\rangle\langle 0|$ and then apply amplitude raising channel $\mathcal{F}$ that maps $|0\rangle\langle 0|$ to the maximally mixed state. The resulting channel is the completely depolarizing channel which is mixed-unitary and hence not extremal.

Amplitude damping channel

Define $\mathcal{E}:L(\mathcal{C}^2)\to L(\mathbb{C}^2)$ by $\mathcal{E}(\rho) = E_0\rho E_0^\dagger+E_1\rho E_1^\dagger$ where

$$ E_0=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad E_1=\begin{bmatrix}0&1\\0&0\end{bmatrix} $$

are linearly independent Kraus operators for $\mathcal{E}$ and

$$ E_0^\dagger E_0=\begin{bmatrix}1&0\\0&0\end{bmatrix}\quad E_0^\dagger E_1=\begin{bmatrix}0&1\\0&0\end{bmatrix}\\ E_1^\dagger E_0=\begin{bmatrix}0&0\\1&0\end{bmatrix}\quad E_1^\dagger E_1=\begin{bmatrix}0&0\\0&1\end{bmatrix} $$

so $\{E_0^\dagger E_0, E_0^\dagger E_1, E_1^\dagger E_0, E_1^\dagger E_1\}$ is also linearly independent and therefore $\mathcal{E}$ is extremal. Moreover, $\mathcal{E}(\rho)=|0\rangle\langle 0|$ for all states $\rho$.

Amplitude raising channel

Next, define $\mathcal{F}:L(\mathcal{C}^2)\to L(\mathbb{C}^2)$ by $\mathcal{F}(\rho) = F_0\rho F_0^\dagger+F_1\rho F_1^\dagger$ where

$$ F_0=\begin{bmatrix}\frac{1}{\sqrt{2}}&0\\0&1\end{bmatrix},\quad F_1=\begin{bmatrix}0&0\\\frac{1}{\sqrt{2}}&0\end{bmatrix} $$

are linearly independent Kraus operators for $\mathcal{F}$ and

$$ F_0^\dagger F_0=\begin{bmatrix}\frac12&0\\0&1\end{bmatrix}\quad F_0^\dagger F_1=\begin{bmatrix}0&0\\\frac{1}{\sqrt{2}}&0\end{bmatrix}\\ F_1^\dagger F_0=\begin{bmatrix}0&\frac{1}{\sqrt{2}}\\0&0\end{bmatrix}\quad F_1^\dagger F_1=\begin{bmatrix}\frac12&0\\0&0\end{bmatrix} $$

so $\{F_0^\dagger F_0, F_0^\dagger F_1, F_1^\dagger F_0, F_1^\dagger F_1\}$ is also linearly independent and therefore $\mathcal{F}$ is extremal.

Depolarizing channel

Now, let $\mathcal{G}:=\mathcal{F}\circ\mathcal{E}$. Since $\mathcal{E}$ is constant, $\mathcal{G}$ is constant and

$$ \mathcal{G}(\rho) = \mathcal{F}(|0\rangle\langle 0|) = \frac{I}{2}. $$

In other words, $\mathcal{G}$ is the completely depolarizing channel which has the following Kraus representation

$$ \mathcal{G}(\rho)=\frac14\rho + \frac14 X\rho X + \frac14 Y\rho Y + \frac14 Z\rho Z $$

(see e.g. exercise $8.17$ on page $379$ in Nielsen & Chuang). Thus, depolarizing channel is mixed-unitary and therefore not extremal.