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6

A necessary and sufficient condition on the unitary $B$ is that its columns all correspond to maximally entangled states. There also does not need to be any relationship between the two unitaries labeled $B$ and $B^{-1}$ in your figure: as long as you start with a maximally entangled state of systems 2 and 3, and then measure systems 1 and 2 with respect ...

5

No, you need to send only one photon (from the pair). The other party could generate entangled pair and send the entangled photon to you. Or it could be the third party that send both of you your photons $-$ prior to actual communication. Even if it's you who generate the entangled pair and share the entangled photon $-$ you could do it way before the time ...

5

No, you can't send more than 2 bits per transmitted qubit. Ultradense Coding would allow FTL Signalling. The basic problem is that, if either teleportation or superdense coding was slightly more efficient, iteratively nesting them inside of each other would allow you to send more encoded qubits than the number of physical qubits you sent. Suppose you could ...

4

An entangled quantum state, where one subsystem is held by each party, and there is no other communication, cannot be used to achieve communication. The two classes of operation that one party could perform on their quantum system are unitaries and measurements. A unitary performed on one quantum subsystem (Alice) does not change the other one (Bob), so ...

3

You cannot directly send information using only entanglement like DaftWullie excellently explained. But here's what you can do Superdense coding: a) pre-share 2 entangled qubits between you and a friend b) then by sending a single qubit you can send 2 classical bits Teleportation: a) pre-share 2 entangled qubits between you and a friend b) when you want to ...

3

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 ...

2

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 ...

2

In my understanding, the key part for entanglement to increase capacity is to have a suboptimal channel. Suppose the input of you channel can take value in the set $X$, and note $G(X)$ the graph where nodes are possible inputs and there is an edge between to inputs if the range of their corresponding output is overlapping, i.e. When going through the channel ...

2

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 ...

1

It is important to note that Step 5 is a classical step. Different protocols exist to correct keys, for instance the Cascade- and Winnow-protocols. With this you reveal bits of information in specific ways, due to which you learn if there are errors and where these are located.

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