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How to prove that for a multi-qubit system a unital channel is not necessarily mixed unitary? This is Problem 8.3 in Nielsen and Chuang. Here's a snippet of the text:

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Shall I need to take two parallel channels to prove this thing, and instead of $\rho_{AB}$ even I will take $I_{AB}$? It's the case of separable. Can someone please help me prove this for a multi-qubit system?

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    $\begingroup$ Can you write out your problem precisely? With my understanding, a mixed unitary channel would be unital. $\endgroup$
    – Rammus
    Oct 3, 2023 at 9:28
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    $\begingroup$ the problem is asking the opposite of what you wrote: prove that unital doesn't imply mixed-unitary. This is also discussed in chapter 4 of Watrous' book where they give an explicit example. $\endgroup$
    – glS
    Oct 3, 2023 at 9:37
  • $\begingroup$ Does this answer your question? Determining whether there exists an equivalent set of unitary Kraus operators $\endgroup$ Oct 3, 2023 at 9:45
  • $\begingroup$ @AdamZalcman it took me a while to find the connection between that question and this one, but eventually I did find: "Also as Adam has answered, qubit channels are mixed unitary if and only if they are unital. This is no longer true in dimension 3 and higher." However this question asked "Can someone please help me how to prove it is not unital for a multi-qubit system." The "if and only if" theorem for single-qubit channels would no longer apply right? $\endgroup$ Oct 7, 2023 at 1:18
  • $\begingroup$ Indeed. The answer there has a counterexample, but no proof. I agree and have retracted my VTC. $\endgroup$ Oct 8, 2023 at 11:34

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Unital channel which is not mixed unitary

Consider the following two-qubit quantum channel $$ \Xi(X)=A_0XA_0^\dagger+A_1XA_1^\dagger\tag1 $$ where $$ A_0=\begin{bmatrix} 1&&&\\ &&\frac{1}{\sqrt2}&\\ &\frac{1}{\sqrt2}&&\\ &&&\frac{1}{\sqrt2}\\ \end{bmatrix}\quad A_1=\begin{bmatrix} 0&&&\\ &\frac{1}{\sqrt2}&&\\ &&-\frac{1}{\sqrt2}&\\ &&&\frac{1}{\sqrt2}\\ \end{bmatrix}.\tag2 $$ We have $\Xi(I)=I$, so $\Xi$ is unital. Moreover, $A_0$ and $A_1$ are linearly independent and $A_0^2$, $A_0A_1$, $A_1A_0$ and $A_1^2$ are also linearly independent. Therefore, by theorem $2.31$ on page $97$ in John Watrous' book The Theory of Quantum Information (see below), the channel $\Xi$ is an extreme point of the convex set of quantum channels. In other words, $\Xi$ is not a non-trivial convex combination of other quantum channels. In particular, $\Xi$ is not a non-trivial convex combination of unitary channels. Finally, $\Xi$ is not unitary since it sends $|01\rangle$ to a mixed state.

Conditions for a channel to be an extreme point

For ease of reference I reproduce the theorem used above:

Theorem $2.31$ (Choi) Let $\mathcal{X}$ and $\mathcal{Y}$ be complex Euclidean spaces, let$^1$ $\Phi\in C(\mathcal{X}, \mathcal{Y})$ be a channel, and let $\{A_a:a\in\Sigma\}\subset L(\mathcal{X},\mathcal{Y})$ be a linearly independent set of operators satisfying$^2$ $$ \Phi(X)=\sum_{a\in\Sigma}A_aXA_a^\dagger\tag{2.174} $$ 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 $$ \{A_b^\dagger A_a:(a,b)\in\Sigma\times\Sigma\}\subset L(\mathcal{X})\tag{2.175} $$ of operators is linearly independent.


$^1$ The symbol $C(\mathcal{X},\mathcal{Y})$ denotes the set of all quantum channels, i.e. completely positive and trace-preserving linear maps $\Phi:L(\mathcal{X})\to L(\mathcal{Y})$, see definition $2.13$ on page $73$.
$^2$ In the book, the adjoint is denoted $A^*$, see page $11$. Here, I am using the more common notation $A^\dagger$.

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