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Even if such a channel is useless, it's still worthwhile to define them for things like quantum communication, to see whether or not you can use your channel to transmit quantum information. So the channels are "useful" in the sense of recognizing what type of channel you have (mathematical formalism should always make it easier to recognize when ...


3

Minimalist formal proof (I'll use $\mu_a\equiv \mu(a)$): $\textrm{(A)}\Rightarrow\textrm{(B)}:$ Let $\Gamma\ge0$. Then, $$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$ which is a separable decomposition, since $\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0$ because it describes ...


2

That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements $\mu_a$, and on obtaining outcome $a$ preparing the state $\sigma_a$. It should be obvious that this breaks any entanglement, since it (destructively) measures the input. (Note that also ...


2

That is indeed some weirdly written exposition with typos, but the result is correct. Let $\Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho)$ and $\Phi_k(\rho)=R_k \text{Tr}(F_k\rho)$. For $\Gamma = \rho_1 \otimes \rho_2$ we have $$ (I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) = $$ $$ = \rho_1\text{Tr}(F_k\...


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