Existence of a perturbed channel that achieves a perturbed output state

Question

Consider a $d$-dimensional maximally entangled state $\vert\phi\rangle = \frac{1}{d}\sum_{i=1}^d\vert i\rangle_A\vert i\rangle_B$. Let $N_{A\rightarrow A'}$ be a quantum channel and consider $\rho_{A'B} = (N\otimes I_B)\vert\phi\rangle\langle\phi\vert$. I am interested in the set of nearby quantum states $S = \{\tilde{\rho}_{A'B}\ |\ \|\tilde{\rho} - \rho\|_1\leq \varepsilon, \tilde{\rho}_B = \rho_B\}$ for some $\varepsilon\in [0,1]$.

For any $\tilde{\rho}\in S$, does there exist a channel $\tilde{N}_{A\rightarrow A'}$ that outputs it given a maximally entangled input? That is $(\tilde{N}\otimes I_B)\vert\phi\rangle\langle\phi\vert = \tilde{\rho}_{A'B}$? If not, what is a good counterexample?

If such a $\tilde{N}$ exists, then is it close to $N$ in diamond distance as a function of $\varepsilon$?

Answer 1

Yes, the channel $\tilde{N}$ necessarily exists.

Notice first that the state $\rho_B$ is the completely mixed state $\mathbb{1}/d$. So, in order for $\tilde{\rho}_{A'B}$ to be contained in $S$, three things must be true:

1. $\tilde{\rho}_{A'B}$ must be positive semidefinite.
2. $\tilde{\rho}_{B} = \mathbb{1}/d$.
3. $\|\rho_{A'B} - \tilde{\rho}_{A'B}\|_1 \leq \varepsilon$.

In general, the state $$ (M \otimes I_B) | \phi \rangle \langle \phi | $$ uniquely determines the mapping $M$ (for any $M$): up to the normalization $1/d$, this is the Choi representation (or Choi-Jamiolkowski representation) of $M$.

Thus, there is a unique map $\tilde{N}$ such that $\tilde{\rho}_{A'B} = (\tilde{N}\otimes I_B)|\phi\rangle\langle \phi|$. The first condition listed above guarantees that $\tilde{N}$ is completely positive and the second condition guarantees that $\tilde{N}$ preserves trace, so $\tilde{N}$ must in fact be a channel.

Concerning the closeness of $\tilde{N}$ to $N$ in diamond distance, the best you can conclude without more information is that $\|\tilde{N} - N\|_{\diamond} \leq d \varepsilon$. See this answer on Theoretical Computer Science Stack Exchange for further details.

Answer 2

1. It is sufficient but not necessary for a channel $N'$ to be $\epsilon-$close in diamond distance with $N$.
2. I think a post-processing noisy channel $E_{A'\to A'}$ can always be found such that $\tilde{N} = (E_{A'}\otimes I)\circ N$ and $(E_{A'}\otimes I_B)\circ N_{A\to A'}\otimes I (\phi_{AB}) = E_{A'}\otimes I_B (\rho_{A' B}) = \tilde{\rho}_{A'B}$.