Rank of mixed-unitary channel

Question

Let $\mathcal{X}$ be a complex Euclidean space, let $n = \text{dim}(\mathcal{X})$, and let $Φ ∈ C(\mathcal{X})$ be a mixed-unitary channel (it means that $\forall X,Φ(X)=\sum_{a}p_aU_aXU^{*}_a$). Prove that there exists a positive integer $m$ satisfying $m ≤ n^4 − 2n^2+2$, a collection of unitary operators ${U_1, . . . , U_m} ⊂ U(\mathcal{X})$, a probability vector $(p_1, . . . , p_m)$ such that $Φ(X) =\sum_{k=1}^mp_kU_kXU^{*}_k$ for all $X$.

This problem is from Proposition 4.9 of https://cs.uwaterloo.ca/~watrous/TQI/TQI.4.pdf and there is a proof in the link. But I don't understand what this means: "The Choi representation of $Ψ_U$ is therefore drawn from an affine subspace of $Herm(\mathcal{X} ⊗ \mathcal{X} )$ having dimension $n^4 − 2n^2 + 1$". Also $\Xi$ defined there is not a bijection, what's the purpose of defining this map?

Also any ideas different from the proof in the link are welcomed.

Answer 1

In Eq. (4.52) Watrous shows that for all $U\in\mathsf U(\mathcal X)$ $$ \Xi(J(\Psi_U))=\begin{pmatrix}{\bf 1}&0\\0&{\bf 1}\end{pmatrix} $$ which is equivalent to $ J(\Psi_U)-\tfrac1n{\bf 1}\otimes{\bf 1}\in\ker\Xi $ as is readily verified. Equivalently, $$ \big\{J(\Psi_U):U\in\mathsf U(\mathcal X)\big\}\in\ker\Xi+\frac1n{\bf 1}\otimes{\bf 1}\subseteq\text{Herm}(\mathcal X\otimes\mathcal X) $$ which together with Eq. (4.51) shows that the Choi representation of $\Psi_U$ is drawn from $\{J(\Psi_U):U\in\mathsf U(\mathcal X)\}$ which is an affine ($+\frac1n{\bf 1}\otimes{\bf 1}$) subspace of $\text{Herm}(\mathcal X\otimes\mathcal X)$ of dimension $\dim(\ker\Xi)=(n^2-1)^2=n^4-2n^2+1$.

The reason Watrous introduces the map $\Xi$ in the first place is that once we know that the $J(\Psi_U)$ --- which by definition of mixed unitaries are the building blocks of all Choi representations $J(\Phi)$ we care about --- it suffices to know the dimension of the space these $J(\Psi_U)$ live in; the reason for this is Carathéodory's theorem which turns the dimension of this subspace into an upper bound on the number of unitaries one needs.

Of course we could use the brute force argument that all $J(\Psi_U)$ live in $\text{Herm}(\mathcal X\otimes\mathcal X)$ so Carathéodory would yield $m\leq\dim(\text{Herm}(\mathcal X\otimes\mathcal X))+1=n^4+1$ where we used that the Hermitian $n\times n$ matrices have (real) dimension $n^2$. With this in mind introducing $\Xi$ helped us to find a better upper bound for the space the unitary channels live in which in turns yielded a better upper bound for $m$.