Prove that $A\preceq B$ implies $A=\Psi(B)$ for some channel $\Psi$

Question

Define $\newcommand{\PP}{\mathbb{P}}\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\tr}{\operatorname{tr}}\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert}\PP_\psi\equiv\ketbra\psi$, and let $\ket\psi,\ket\phi$ be two bipartite states such that $\tr_2(\PP_\psi)\prec\tr_2(\PP_\phi)$. Here, $A\prec B$ with $A,B$ positive operators means that the vector of eigenvalues of $A$ is majorised by that of $B$: $A\preceq B\Longleftrightarrow\lambda(A)\preceq\lambda(B)$.

A step to prove Nielsen's theorem, used in the proof of the theorem given here (pdf alert) is that $\tr_2(\PP_\psi)\prec\tr_2(\PP_\phi)$ implies $\tr_2(\PP_\psi)=\Psi(\tr_2(\PP_\phi))$ for some mixed unitary channel $\Psi$. More precisely, it implies that $\tr_2(\PP_\psi)=\Psi( W\tr_2(\PP_\phi)W^\dagger)$ for some mixed unitary channel $\Psi$ and isometry $W$ (though these two statements seem pretty much equivalent to me).

To show this, an important observation seems to be the fact that, introducing the operators $X,Y$ with components $X_{ij}=\psi_{ij}, Y_{ij}=\phi_{ij}$ (that is, $\ket\psi= \operatorname{vec}(X)$ and $\ket\phi= \operatorname{vec}(Y)$), we have $$\tr_2(\PP_\psi) = XX^\dagger,\qquad \tr_2(\PP_\phi) = YY^\dagger.$$ Suitably defining the underlying vector spaces, we can always assume $XX^\dagger ,YY^\dagger >0$. Moreover, $XX^\dagger\prec YY^\dagger$ implies $\operatorname{rank}(XX^\dagger)\ge\operatorname{rank}(YY^\dagger)$.

Why does this imply that the existence of a mixed unitary channel $\Phi$ and isometry $W$ such that $XX^\dagger = \Psi(WYY^\dagger W^\dagger)$? The reason is probably trivial but I'm not seeing it right now.

Answer 1

Let $\rho_{d}, \sigma_{d}$ be the (simultaneously diagonal) density matrices whose eigenvalues are $\{ p_{j} \}, \{ q_{j} \}$, respectively (represented as probability vectors below). Then, if $\vec{p} \succ \vec{q}$, the following sequence of arguments can be observed:

1. There exists a bistochastic matrix $M$ such that $M \vec{p} = \vec{q}$ (basic result of majorization theory, see Marshall and Olkin, for example.)
2. Using Birkhoff's theorem, the bistochastic can be written as a convex combination of permutations: $M = \sum\limits_{j} r_{j} P_{j}$.
3. $M$ can be ``quantized'' into a (mixed unitary) CPTP map, $M \mapsto \mathcal{M} = \sum\limits_{j} r_{j} \mathcal{U}_{P_{j}}$, where $\mathcal{U}_{P_{j}}$ is the unitary superoperator, defined as $\mathcal{U}_{P_{j}}(\cdot) = P_{j} (\cdot) P_{j}^{\dagger}$. Recall that permutations have a unitary representation.
4. The action of $\mathcal{M}$ is to transform $\rho_{d} \mapsto \sigma_{d}$.

Why can we start from simultaneously diagonal states $\rho_{d}, \sigma_{d}$? Hint: the partial trace.

In several quantum resource theories the state transformation reduces to ``classical majorization'', i.e., majorization of vectors (as opposed to say matrix majorization), for example, resource theory of coherence, non-uniformity, etc.