# Existence of a perturbed channel that achieves a perturbed output state

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 $$\varepsilon$$?

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.

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 $$(E_{A'}\otimes I_B)\circ N_{A\to A'}\otimes I (\phi_{AB}) = E_{A'}\otimes I_B (\rho_{A' B}) = \tilde{\rho}_{A'B}$$.