# Unital channel which is not mixed unitary

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:

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?

• Can you write out your problem precisely? With my understanding, a mixed unitary channel would be unital. Oct 3, 2023 at 9:28
• 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.
– glS
Oct 3, 2023 at 9:37
• Does this answer your question? Determining whether there exists an equivalent set of unitary Kraus operators Oct 3, 2023 at 9:45
• @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? Oct 7, 2023 at 1:18
• Indeed. The answer there has a counterexample, but no proof. I agree and have retracted my VTC. Oct 8, 2023 at 11:34

## 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 $$A^\dagger$$.