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.

  • $\begingroup$ 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. $\endgroup$ Jul 22, 2020 at 18:57
  • $\begingroup$ 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.) $\endgroup$ Jul 22, 2020 at 18:59
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    $\begingroup$ @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 $\endgroup$
    – glS
    Jul 22, 2020 at 19:02
  • $\begingroup$ 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.) $\endgroup$ Jul 22, 2020 at 19:03
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    $\begingroup$ 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. $\endgroup$ Jul 22, 2020 at 20:01

1 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.

  • $\begingroup$ @gIS Does this answer your question? $\endgroup$ Jul 23, 2020 at 20:22
  • $\begingroup$ 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 $\endgroup$
    – glS
    Jul 23, 2020 at 20:29
  • $\begingroup$ 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 $\endgroup$
    – glS
    Jul 23, 2020 at 20:38
  • $\begingroup$ Could you explain the hint again? $\endgroup$
    – Cain
    Jul 7 at 9:23

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