The superdense coding protocol requires that A and B share an entangled state. Why is that? That is, why does the shared state have to be entangled? Does the protocol (implicitly) use the fact that the state is entangled?
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Superdense coding is a protocol used to transmit two classical bits of information from one party to the other by using only a single qubit and no classical communication between them. It is a corollary of Holevo's theorem that 1 qubit can not hold more than 1 bit of information. A communication protocol which demonstrates the transmission of 1 classical bit from (say) Alice to Bob via a quantum channel would be that Alice prepares a qubit in either of the states |0> or |1> and transmits it to Bob. Bob then measures it in the computational basis to recover tha classical bit. This protocol only conveys 1 classical bit while transmitting 1 quantum bit (so no entanglement is involved). However, if the two parties already have a shared nonlocal resource (the 1 ebit), then it is possible to transmit 2 classical bits via superdense coding, at the cost of consuming 1 ebit resource. (The shared classical information would be private as well due to maximum entanglement in shared qubits). Once they have a shared ebit, Alice only needs to transmit her (single) qubit to transmit 2 classical bits (apart from applying local operations). The distinguishability of the Bell states upon a measurement in the Bell basis by Bob is where the entanglement is being used to achieve this. The increase in the number of bits conveyed per number of qubits transmitted is increased, but only at the cost of Alice and Bob having shared an entangled resource in the past.
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$\begingroup$ Can we not distinguish the Bell states because they are orthogonal to one another? I think I'm having a problem understanding entanglement. What exactly does it mean to 'share' the ebit? So changes by Alice on her qubit change the ebit? Because they are entangled? And so Bob needs Alice's qubit to reconstruct this changed ebit? Is this the idea? $\endgroup$ Oct 13, 2020 at 20:48
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$\begingroup$ 1. The 4 Bell states are orthogonal, that is why they are distinguishable. The thing with superdense coding is A.)If you insist on transmitting 2 classical bits of information without using the shared entanglement, you have to transmit 2 qubits via the quantum channel. The 4 possible messages formed by 2 bits can be encoded in the 4 basis states, which are distinguishable. B.) When using shared ebit, Alice will necessarily have to transmit her qubit to Bob, otherwise Bob can't decode the message (try to measure just 1 qubit out of 2 completely entangled qubits-you get 0/1 with 50% probability. $\endgroup$– bhqOct 15, 2020 at 17:50
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$\begingroup$ 2. Sharing the ebit means the following - at the beginning of the protocol, a third participant, Charlie, prepares 2 qubits in a Bell state. He gives 1 qubit to Alice & 1 to Bob - now Alice & Bob are said to share entanglement. Once the 2 qubits are entangled, you have to stop thinking of them as individual entities with independent states - all you get to say is: the two qubits are in this Bell state. (this is the non intuitive part about entangled states). So, when Alice applies a local operation on her qubit of the entangled pair, it changes the state of the entangled pair of qubits. $\endgroup$– bhqOct 15, 2020 at 17:59
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