Why are entanglement breaking channels, defined as $\Phi(\rho)=\sum_a \operatorname{Tr}(\mu(a)\rho)\sigma_a$, entanglement breaking?

Question

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)?

Answer 1

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 $\mathrm{tr}(X)=\mathrm{tr}(X^T)$.

Answer 2

There are some 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.

Answer 3

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 $\mu_a$.)

Answer 4

Here's another approach:

1. Show that a channel $\Phi$ "breaks entanglement" (i.e. satisfies (B) in the question) iff its Choi representation is separable.

   The Choi representation is here defined as $J(\Phi)\equiv (\Phi\otimes\operatorname{Id})(\Omega)$, with $\Omega\equiv \frac1{d}\sum_{ij}|ii\rangle\!\langle jj|$ the maximally entangled state.

   That (B) implies separable $J(\Phi)$ is immediate from the definition of Choi. On the other hand, we can write the action of $\Phi\otimes \operatorname{Id}$ on a generic bipartite pure state $|\chi\rangle$ as $$(\Phi\otimes \operatorname{Id})\mathbb{P}_\chi = (I\otimes \sqrt D) J(\Phi) (I\otimes \sqrt D), \quad\mathbb{P}_\chi\equiv|\chi\rangle\!\langle\chi|, $$ where $|\chi\rangle=\sum_i \sqrt{d_i} |i,i\rangle$, $d_i>0$, is a Schmidt decomposition for $|\chi\rangle$, and $D_{ij}=\delta_{ij} d_i$ (a similar argument is used here to show that $n$-positivity is equivalent to complete positivity). It follows that if $J(\Phi)$ is separable then so is $(\Phi\otimes \operatorname{Id})\mathbb{P}_\chi$, for any $|\chi\rangle$. This, in turn, implies that $(\Phi\otimes\operatorname{Id})\Gamma$ must be separable for any state $\Gamma$, as any state is a convex combination of pure ones.

2. Show that $\Phi$ EB (as per (A)) implies separable Choi.

   This is an easy consequence of the Choi having the form $$J(\Phi) = \frac1d\sum_a \sigma_a\otimes \,\bar\mu_a.$$

3. Show that any channel $\Phi$ with separable Choi is EB, that is, can be cast in the form (A).

   For this we use the following general property of Chois: a map has the form $\Phi(X)=A \langle B,X\rangle$ iff its Choi is $J(\Phi)=\frac1d A\otimes\bar B$. It follows for us that if the Choi is separable, i.e. it's writable as $J(\Phi)=\frac1d\sum_a A_a\otimes \bar B_a$ for some positive semidefinite $A_a,B_a$, then the channel reads $$\Phi(\rho) = \sum_a A_a \langle B_a,\rho\rangle.$$ That $\{B_a\}_a$ must be a POVM and $A_a$ must be states then follows from the requirement of $\Phi$ being CPTP.