New answers tagged

2

$$I(A:B)=S(A)+S(B)-S(AB)$$ $$J(A_{\{\Pi_{i}\}}:B)=S(A_{\{\Pi_{i}\}})+S(B)-S(A_{\{\Pi_{i}\}}B)$$ $$I(A:B)-J(A_{\{\Pi_{i}\}}:B)=S(A)-S(AB)-S(A_{\{\Pi_{i}\}})+S(A_{\{\Pi_{i}\}}B)$$ Since $$\rho = \sum_j p_j \pi_j\otimes \rho_j,$$ it is invariant to the measurements performed on the subsystem A, after the results are forgotten, so $$S(A)=S(A_{\{\Pi_{i}\}})$$ and ...


Top 50 recent answers are included