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Mathematically, this has nothing to do with the positivity of $E$. It doesn't really have anything to do with $E$ at all - it's a property of the Bell states themselves (you've probably not got there yet, but they have the same reduced density matrices). I presume the reason for specifying the positivity of $E$ in the question is to help you make the ...


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This happens for any maximally entangled state $|\Psi\rangle$ and operator $E$. Indeed, a maximally entangled state is, by definition, one whose partial trace is the maximally mixed one. Writing $|\Psi\rangle\equiv\sum_{ij}\psi_{ij}|ij\rangle$, this means that $\sum_i \psi_{ij}\bar\psi_{kj}=\delta_{ik}/D$, with $D$ the dimension of each space ($D=2$ in your ...


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I will try to succinctly answer your first question given that I possess little knowledge regarding entanglement assistance. Shannon's capacity theorem (the noisy channel coding theorem) states that for any code $\mathbf{C}$ of code rate $R \leq C$, where $C$ represents the channel capacity, an encoding and decoding scheme of rate $R$ with a probability of ...


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