# Action of a CPTP map on Identity

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 $$\sum_i K_i K_i^+ \leq \mathbb{I}$$?

• 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. Jul 4, 2023 at 15:53
• Of course, thought it best to write it explicitly.
– Cain
Jul 4, 2023 at 16:56
• 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. Jul 7, 2023 at 19:26

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 $$\sum_i K_iK_i^\dagger = 2|{0}\rangle\langle{0}|\nleq\mathbb 1$$.

• As I understand it, unless we have more information about a specific map, we are not able to say anything better than this?
– Cain
Jul 5, 2023 at 6:45
• 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 $\Phi$. Jul 5, 2023 at 14:22

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.

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