Define an entanglement breaking channel $\Phi$ as a channel (CPTP map) of the form $$\Phi(\rho) = \sum_a \operatorname{Tr}(\mu(a)\rho) \sigma_a\tag A$$ for some POVM $\{\mu(a)\}_a$ and states $\sigma_a$.

It is mentioned e.g. in (Horodecki, Shor, Ruskai 2003) that $\Phi$ is entanglement breaking iff it "breaks entanglement", that is, is such that $$(\Phi\otimes I)\Gamma \quad\text{ is separable for every state } \Gamma.\tag B$$

This equivalence is proved, I think, in pages 5 and 6 of the above reference, but I can't quite follow the exposition there. In particular, the proof that (A) implies (B) seems to rely on expressing the action of $\Phi\otimes I$ on $\Gamma$ via a partial trace involving some operators $E_k$ which however are not defined (there might be a typo somewhere in the text, I'm not sure).

What are good ways to prove the equivalence between (A) and (B)?

  • $\begingroup$ Presumably, the E_k are the mu_a. (A)=>(B) is completely trivial. The backward direction is probably easiest via Choi-Jamiolkowski (the mu_a and sigma_a are simply the states in the separable decomposition of the Choi state). $\endgroup$ – Norbert Schuch Mar 19 at 13:15

Minimalist formal proof (I'll use $\mu_a\equiv \mu(a)$):

$\textrm{(A)}\Rightarrow\textrm{(B)}:$ Let $\Gamma\ge0$. Then, $$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$ which is a separable decomposition, since $\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0$ because it describes the action of the CP map $\rho\mapsto\mathrm{tr}(\mu_a\rho)$ (measurement conditioned on outcome) on the $A$ system.

$\textrm{(B)}\Rightarrow\textrm{(A)}:$ Let $\chi_{AB}=(\Phi_A\otimes I_B)(\Omega_{AB})=\sum \tilde\sigma_a\otimes \tilde\mu_a$ be the Choi state of $\Phi$ (with $\Omega$ the maximally entangled state). Then, the map $\Phi$ can be obtained from $\chi_{AB}$ as \begin{align*} \Phi(\rho) &= \mathrm{tr}_B(\chi_{AB}(I\otimes\rho^T))\\ &=\sum \mathrm{tr}_B((\tilde\sigma_a\otimes\tilde\mu_a)(I\otimes\rho^T))\\ &=\sum \tilde\sigma_A\,\mathrm{tr}(\tilde\mu_a\rho^T)\\ &=\sum \tilde\sigma_A\,\mathrm{tr}(\tilde\mu_a^T\rho) \end{align*} where the last equality follows from $\mathrm{tr}(X)=\mathrm{tr}(X^T)$.

  • $\begingroup$ Intuitively, the transpose is related to the different "direction" in which the objects act: $\Phi$ acts on a state, while $\tilde\mu_a$ is a state, so for it to act on a state the indices have to be flipped. (Formally, this amounts to putting $\tilde\mu_a$ and $\rho$ next to each other and projecting onto the ME state.) $\endgroup$ – Norbert Schuch Mar 22 at 0:27
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    $\begingroup$ Nice, this kind of formalism is acceptable (for me), while this derivation is simpler conceptually. $\endgroup$ – Danylo Y Mar 22 at 13:43

That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements $\mu_a$, and on obtaining outcome $a$ preparing the state $\sigma_a$. It should be obvious that this breaks any entanglement, since it (destructively) measures the input.

(Note that also proving this direction is a one-liner - the output will be $\mu_a$ times the partial trace of the input with $\mu_a$.)

  • $\begingroup$ I don't disagree with the result being physically intuitive, which is part of the reason why I'd expect it to be reachable with a "proportionally simple" mathematical proof, which the one given in the reference isn't, imo. But yea, if I'm deciphering your last sentence correctly, that's certainly a better way to look at it. I think you are just noting that $(\Phi\otimes I)\Gamma=\sum_a \sigma_a \otimes \mathrm{Tr}_1[(\mu(a)\otimes I)\Gamma]$, yes? This reflects the physical intuition nicely. $\endgroup$ – glS Mar 21 at 16:55
  • $\begingroup$ (which looking back at the paper is the same thing they were doing, though I was confused by their notation). For the other direction, from the other comment, you take $\Gamma$ as the maxent state, thus getting the Choi, from which you can infer the structure of the channel itself. I guess the physical observation here is that how the extension acts on maximally entangled states determines the action of the channel itself (which is probably the intuition underlying the usefulness of the Choi in general). $\endgroup$ – glS Mar 21 at 16:59
  • $\begingroup$ @glS Indeed, that's what I meant. (You just have to prove that this is positive, but since it is the residual state after applying a CP map $\mathrm{tr}(\mu(a))\,\cdot\,)$ to half the system, this is clear.) For the converse, if you know how the backwards direction of the Choi isomorphism works, namely $\mathcal E(\rho) = \mathrm{tr}_2 (\sigma(I\otimes \rho^T))$, with $\sigma$ the Choi state), it is also immediate. -- I was in fact thinking of writing a minimalist formal answer. Would that be of interest? $\endgroup$ – Norbert Schuch Mar 21 at 17:01
  • $\begingroup$ I think so!$\,\,\,$ $\endgroup$ – glS Mar 21 at 17:23
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    $\begingroup$ Done! (And I learned something doing it!) $\endgroup$ – Norbert Schuch Mar 21 at 18:26

That is indeed some weirdly written exposition with typos, but the result is correct.

Let $\Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho)$ and $\Phi_k(\rho)=R_k \text{Tr}(F_k\rho)$.

For $\Gamma = \rho_1 \otimes \rho_2$ we have $$ (I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) = $$ $$ = \rho_1\text{Tr}(F_k\rho_2) \otimes R_k = \text{Tr}_2(\rho_1\otimes F_k\rho_2) \otimes R_k = $$ $$ = \text{Tr}_2(I\otimes F_k \cdot \Gamma) \otimes R_k = \text{Tr}_2(I\otimes \sqrt{ F_k} \cdot \Gamma \cdot I\otimes \sqrt{F_k}) \otimes R_k. $$ By linearity, it's true for every $\Gamma$, and also by linearity we can write that $$ (I \otimes \Phi)(\Gamma) = \sum_k \text{Tr}_2(I\otimes\sqrt{ F_k} \cdot \Gamma \cdot I\otimes\sqrt{ F_k}) \otimes R_k $$ This proves (A) => (B) since for density matrix $\Gamma$ we have $\text{Tr}_2(I\otimes\sqrt{ F_k} \cdot \Gamma \cdot I\otimes \sqrt{ F_k}) =: \gamma_k Q_k$ where $\gamma_k>0$ and $Q_k$ is a density matrix (they also have typo in the definition of $Q_k$).

To prove (B) => (A) we take $\Gamma = |\beta\rangle\langle\beta|$, where $|\beta\rangle = \frac{1}{\sqrt{d}}\sum_i |i\rangle|i\rangle$ and use the separability of $(I \otimes \Phi )(\Gamma)$. That is, If $$(I \otimes \Phi )(|\beta\rangle\langle\beta|)=\sum_n p_n |v_n\rangle\langle v_n| \otimes |w_n\rangle\langle w_n|$$ then it can be showed that $$\Phi(\rho) = \Omega(\rho) := d\sum_n |w_n\rangle\langle w_n| \text{Tr}\big(\rho p_n \big(|v_n\rangle\langle v_n|\big)^T\big).$$ To prove $\Phi = \Omega$ it's enough to show that $(I \otimes \Phi )(|\beta\rangle\langle\beta|) = (I \otimes \Omega )(|\beta\rangle\langle\beta|)$.

This is the (C) => (A) implication of Theorem 4 in the paper (that has a minor mistake).

  • $\begingroup$ What's the point of taking square roots of the F_k? Also, the "To prove Phi=Omega" - isn't that just the backwards map of Choi-Jamiolkowski? $\endgroup$ – Norbert Schuch Mar 19 at 13:17
  • $\begingroup$ @NorbertSchuch It's then easier to see why $I \otimes F_k \cdot \Gamma$ is positive semidefinite (if $\Gamma$ is), also it's aligned with the presentation from the paper. It might be unnecessary and obvious at all from physical intuition, but I prefer to support that intuition by precise math. The state $(I \otimes \Phi)(|\beta\rangle\langle\beta|)$ is the Choi's matrix of $\Phi$, so of course $\Phi$ is an image of the backwards map. $\endgroup$ – Danylo Y Mar 19 at 13:46
  • $\begingroup$ I agree - I just felt it is not required if one wants the shortest possible proof (different people have different primary interests ;) ): After all, it is the post-measurement state on the 1-system, so it must be positive. Regarding the 2nd point, I was just wondering why you said "To prove Phi=Omega" given it is the backwards Choi state. Or did you not want to use that? (In some sense, you anyway use Choi, because you use that the channel is characterized by its action on the ME state.) $\endgroup$ – Norbert Schuch Mar 19 at 15:17
  • $\begingroup$ You know this from physical intuition that $Q_k$ is a post-measurement state (so it must be positive), which is disentangled from the preparation $R_k$. It's good to know this explanation, but this can't be used as a rigorous proof, in my opinion (which is biased because I'm mathematician :)). As for 2nd point, the introduction of a new variable $\Omega$ is indeed unnecessary, but I just followed the paper here. In essence, we just need to show that the backwards image of a separable Choi matrix is a channel that has Holevo form. $\endgroup$ – Danylo Y Mar 19 at 15:53
  • $\begingroup$ If you know that the "physical intuition" is not just "intuition", but you have been proving it when learning QI (where such things should be proven), then it is a formal argument. It is only not a formal argument if you never proved it and only base it on physical intuition. It is like I say "it is intuitive the projection of one vector onto another unit vector is shorter than the original vector" and you tell me I cannot use Cauchy-Schwarz if I am led by intuition. Many mathematicians are led by intuition: It tells you which arguments to use! $\endgroup$ – Norbert Schuch Mar 19 at 16:53

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