The set of CPTP maps is convex, therefore, it is enough to perform the needed optimizations over the set of extreme points. Is there any way of characterizing the said extreme points that would lend itself to optimization in general?

As I write this, it occurs to me that one possibility is that there can be infinite vertices (as in the case of the set of separable states), whereby optimizing over the vertices is not particularly simple. Nevertheless, a characterization of the vertices will make matters simpler.

  • 1
    $\begingroup$ There is a characterization via Kraus operators, Thm.2.31 in Watrous "The Theory of Quantum Information" $\endgroup$ Nov 23, 2023 at 17:25
  • 1
    $\begingroup$ By the Choi-Jamiolkowski isomorphism, this is equivalent to the extreme points in the set of corresponding (TP) Choi states. $\endgroup$ Nov 23, 2023 at 17:38
  • $\begingroup$ I know about the Theorem 2.31, but the condition on the Kraus operators does not lend itself nicely to optimization, due to difficulty in generation of such maps. Generating Pure Choi states on the other hand seems simpler to handle. $\endgroup$
    – Cain
    Nov 27, 2023 at 12:26

1 Answer 1



The set $C(\mathcal{X},\mathcal{Y})$ of all channels$^{1,2}$ $\Phi:L(\mathcal{X})\to L(\mathcal{Y})$ has uncountably infinitely many extreme points. To see this, first note that every pure state is an extreme point of the set of density matrices. Consequently, every channel that sends pure states to pure states is necessarily an extreme point of $C(\mathcal{X},\mathcal{Y})$. In particular, every unitary channel is an extreme point of $C(\mathcal{X},\mathcal{Y})$.


Necessary and sufficient conditions for a channel to be an extreme point of the set$^4$ $C(\mathcal{X},\mathcal{Y})$ are provided by theorem $2.31$ on page $97$ in John Watrous' book The Theory of Quantum Information.

Theorem $2.31$ (Choi) Let $\mathcal{X}$ and $\mathcal{Y}$ be complex Euclidean spaces, let $\Phi\in C(\mathcal{X}, \mathcal{Y})$ be a channel, and let $\{A_a:a\in\Sigma\}\subset L(\mathcal{X},\mathcal{Y})$ be a linearly independent set of operators satisfying$^5$ $$ \Phi(X)=\sum_{a\in\Sigma}A_aXA_a^*\tag{2.174} $$ for all $X\in L(\mathcal{X})$. The channel $\Phi$ is an extreme point of the set $C(\mathcal{X}, \mathcal{Y})$ if and only if the collection $$ \{A_b^*A_a:(a,b)\in\Sigma\times\Sigma\}\subset L(\mathcal{X})\tag{2.175} $$ of operators is linearly independent.

Using this theorem, we can find more extreme points of $C(\mathcal{X}, \mathcal{Y})$, such as the amplitude damping channel and this unital channel.

$^1$ A channel is a completely positive and trace-preserving linear map.
$^2$ $\mathcal{X}$ and $\mathcal{Y}$ denote the input and output Hilbert spaces$^3$ respectively.
$^3$ In quantum information science we almost always deal with finite-dimensional vector spaces. Every finite-dimensional vector space over a complete field is complete, so using the definition of Hilbert space is strictly speaking an overkill. John Watrous uses the term "complex Euclidean space" instead.
$^4$ C.f. definition $2.13$ on page $73$.
$^5$ Here, $A^*$ denotes the adjoint of $A$, see page $11$.

  • $\begingroup$ "Consequently, every channel that sends pure states to pure states is necessarily an extreme point" -- Why would that follow from the statement before? $\endgroup$ Nov 24, 2023 at 17:19
  • 1
    $\begingroup$ @NorbertSchuch Statement before implies that if pure state $\rho$ is a linear combination of $\rho_1$ and $\rho_2$, then $\rho_1=\rho_2$. Fix pure input $\sigma$ for channel $\mathcal{E}$ and suppose $\mathcal{E}=p\mathcal{A}+q\mathcal{B}$ for some channels $\mathcal{A}$ and $\mathcal{B}$. Then pure state $\mathcal{E}(\sigma)$ is a linear combination of $\mathcal{A}(\sigma)$ and $\mathcal{B}(\sigma)$. Therefore, $\mathcal{A}(\sigma)=\mathcal{B}(\sigma)$. The equality holds for all pure $\sigma$. But pure states contain a basis, so $\mathcal{A}=\mathcal{B}$. Therefore, $\mathcal{E}$ is extreme. $\endgroup$ Nov 24, 2023 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.