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If a map between positive operators $\Phi: X \rightarrow Y$ is also completely positive, it is true that $\Phi\otimes I_A$ is also a positive map for any choice of ancilla operator space $A$.

That is, for any positive semidefinite operator $\rho \in X$, my map gives me a positive semidefinite operator $\Phi(\rho) = \rho' \in Y$. But my state could be entangled i.e. I could have had $\rho = Tr_A(\sigma)$ for some positive semidefinite $\sigma \in X\otimes A$. In this case, the map still works $(\Phi\otimes I_A) \sigma_{XA} = \sigma'_{XA}$ and gives a positive semidefinite output.

My question is regarding the dimension of $A$. I have seen proofs only consider the dimension of $A$ to be equal to the dimension of $X$. Why is it not required to consider a larger ancilla?

EDIT: To see the proof of what is written in the accepted answer, see https://en.wikipedia.org/wiki/Choi%27s_theorem_on_completely_positive_maps

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I'm not sure I understand your question 100%, but:

Given a channel $\Phi$, the Choi state $$ \sigma_\Phi=(\Phi\otimes I)(|\Omega\rangle\langle\Omega|) $$ with $|\Omega\rangle=\tfrac{1}{\sqrt{d}}\sum_{i=1}^d|i,i\rangle$ the maximally entangled state, contains all the information about the channel $\Phi$ (i.e., given $\sigma_\Phi$, we can reconstruct $\Phi$). In particular, if $\sigma_\Phi$ is positive semidefinite, this implies that $\Phi$ is completely positive. Thus, it is sufficient to choose the dimension of the ancilla equal to the dimension of the system, and it is sufficient to test positivity on a single state on that system+ancilla (e.g. the maximally entangled state $|\Omega$, but in fact any pure state with full Schmidt rank will do).

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  • $\begingroup$ Thank you. I think you answered my question (I didn't think about the Choi state at all) but I think for completeness, I will put this link here en.wikipedia.org/wiki/…. It has an explicit proof about why n-positive $\Phi$ implies a completely positive $\Phi$. $\endgroup$ Commented Dec 20, 2018 at 14:53
  • $\begingroup$ ... which follows exactly the lines of my answer. $\endgroup$ Commented Dec 20, 2018 at 15:46

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