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Suppose we have a CPTP map $\Phi(\rho)=\sum_i K_i \rho K_i^+$, such that, $\sum_i K_i^+K_i=\mathbb{I}$.

In case the map preserves Identity, is unital, then we immediately have $\sum_i K_i K_i^+ =\mathbb{I}$. What can we say about the sum $\sum_i K_i K_i^+$ when the map is not unital? Can we write relations/inequalities like $\sum_i K_i K_i^+ \leq \mathbb{I}$?

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  • $\begingroup$ Note that saying $\Phi(X)=\sum_i K_i X K_i^*$ is CPTP is equivalent to the condition that $\sum_i K_i^*K_i=I$ so its not necessary to include it as an extra assumption. $\endgroup$
    – Condo
    Jul 4, 2023 at 15:53
  • $\begingroup$ Of course, thought it best to write it explicitly. $\endgroup$
    – Cain
    Jul 4, 2023 at 16:56
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    $\begingroup$ Please don't change the question (in the sense of asking a different question) after answers have been given. This invalidates existing answers. Ask a new question, referring to this one. $\endgroup$ Jul 7, 2023 at 19:26

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Let $d$ be the dimension of the system. Since $\sum_i K_iK_i^\dagger = \Phi(\mathbb 1) = d \Phi(\mathbb 1/d)$, it can be $d$ times any quantum state (by, e.g., setting $\Phi$ as a replacement channel). In other words, it can be any trace-$d$ positive semi-definite operator.

Concretely, for a single qubit, by setting $K_0 = |{0}\rangle\langle{0}|$, $K_1 = |{0}\rangle\langle{1}|$, we have $\sum_i K_iK_i^\dagger = 2|{0}\rangle\langle{0}|\nleq\mathbb 1$.

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  • $\begingroup$ As I understand it, unless we have more information about a specific map, we are not able to say anything better than this? $\endgroup$
    – Cain
    Jul 5, 2023 at 6:45
  • $\begingroup$ Yes, as this shows $\sum_i K_i K_i^\dagger$ must be a trace-$d$ positive semi-definite operator (since $\Phi$ is CPTP) and can be any such operator. We are not able to say anything better without more restriction of $\Phi$. $\endgroup$ Jul 5, 2023 at 14:22
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One can show concretely that a $\tfrac1d\Phi(I)$ can be any density matrix $\sigma$. To this end, simply choose $$ \Phi(\rho) = \mathrm{tr}(rho)\,\sigma\ . $$ It can be straightforwardly checked it is CP, e.g., by computing the Choi state, or just by realizing it describes a simple physical process: Throw away the input and prepare a state $\sigma$ as the output.

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  • $\begingroup$ But if there are two parties and one was modelling LOCC, such a map would not arise, would it? $\endgroup$
    – Cain
    Jul 6, 2023 at 7:42
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    $\begingroup$ @Cain You could prepare any separable state with LOCC. $\endgroup$
    – Rammus
    Jul 6, 2023 at 9:44

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