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

Define $$\newcommand{\PP}{\mathbb{P}}\newcommand{\ket}{\lvert #1\rangle}\newcommand{\tr}{\operatorname{tr}}\newcommand{\ketbra}{\lvert #1\rangle\!\langle #1\rvert}\PP_\psi\equiv\ketbra\psi$$, 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.

• Isn't this standard textbook material (I certainly explain the proof in my lecture)? In any case, as far as I remember (would have to check) you need to use the corresponding classical result, namely that majorization for probability distributions implies (even is equivalent to) the existence of a stochastic map (which in turn is a convex combination of permutations (Birkhoff's theorem), from which you construct the channel). But this is purely out of my memory. – Norbert Schuch Jul 22 at 18:57
• quantuminfo.physik.rwth-aachen.de/cms/Quantuminfo/Studium/…, Lecture 8, the theorem on pg 69? (The whole story starts on pg. 65, and takes 6 pages. It is handwritten notes, so it's not that much material.) – Norbert Schuch Jul 22 at 18:59
• @NorbertSchuch I guess it might be in some places =)? I was suspecting it was related to those results, but I'm not that well-versed with majorization-related things. I'll have a look at the lecture, thanks – glS Jul 22 at 19:02
• I think I basically took it from the review by Nielsen and Vidal I link next to the lecture: michaelnielsen.org/papers/majorization_review.pdf (But it has been a while, so I might also have taken material from somewhere else.) – Norbert Schuch Jul 22 at 19:03
• Let me also suggest section 6.2.1 of the book cs.uwaterloo.ca/~watrous/TQI as an alternative to the notes linked in the question. – John Watrous Jul 22 at 20:01

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.

• @gIS Does this answer your question? – keisuke.akira Jul 23 at 20:22
• probably, thanks for the answer! I just need to find a bit of time to check and understand properly the various steps. Regardless, I generally prefer to not accept answers in the first few days after I ask them, as it encourages other people to also give other answers. Don't worry, I always accept received good answers.. eventually – glS Jul 23 at 20:29
• I don't get your "hint" though. The partial trace doesn't necessarily give simultaneously diagonal matrices. But you probably don't need them to be: if I understand this correctly, $\mathcal M$ gives you a matrix with the correct eigenvalues. You then just need to apply a unitary map to change the basis into the correct one – glS Jul 23 at 20:38