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 $\Phi_{1} = \Phi_{2} = \Phi$.


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.

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    $\begingroup$ What about considering a unitary channel $U(\cdot)U^\dagger$ and the inverse channel $U^\dagger(\cdot)U$? The composition would be the identity map, which is not extremal? $\endgroup$ Commented Dec 13, 2021 at 14:37
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    $\begingroup$ @user1936752 Why is the identity channel not extremal? It is a unitary channel and therefore extremal (see for instance example $2.32$ from Watrous' book)? $\Phi_{1}$ and $\Phi_{2}$ are supposed to be channels, not just any map. Or am I missing something? $\endgroup$
    – JSdJ
    Commented Dec 13, 2021 at 14:52
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    $\begingroup$ Ah good point, ignore my comment. $\endgroup$ Commented Dec 13, 2021 at 14:53
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    $\begingroup$ Nice question. I believe the anti-symmetric Werner-Holevo channel in three dimensions provides a counter-example. It is extremal (as is shown in Example 4.3 in my book), but if you compose it with itself you will get 9 linearly independent Kraus operators, which is too many to be able to satisfy Choi's theorem (Theorem 2.31). $\endgroup$ Commented Dec 13, 2021 at 17:56
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    $\begingroup$ Thank you @JohnWatrous, that is also a nice counterexample (which, in addition, also made me think about the maximum number of lin. ind. Kraus operators an extremal channel could have). In addition, thank you for providing these kind of 'answered-by-the-author' comments; I feel they add a lot to the site! $\endgroup$
    – JSdJ
    Commented Dec 15, 2021 at 20:16

1 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.

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    $\begingroup$ Thanks Adam, that's the concise clean counterexample I was hoping for! $\endgroup$
    – JSdJ
    Commented Dec 13, 2021 at 18:15
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    $\begingroup$ You're welcome! Thank you for a nice question! :-) $\endgroup$ Commented Dec 13, 2021 at 18:19

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