12 votes
Accepted

The process for transferring qubits between locations

You are totally right in your assumption about transporting qubits from Alice to Bob implies something physical. Usually problems/situations that have this setup of a transmission between two parties ...
11 votes

What is a flying qubit?

Preliminary - The DiVincenzo criteria for a 'normal' quantum computer The DiVincenzo criteria, as originally proposed by DiVincenzo, are $5$ criteria that he originally proposed in his seminal 2000 ...
  • 4,898
11 votes

Does entanglement allow enhanced communication efficiency?

First, about teleportation, you say that you think quantum communication takes place in the protocol, but it doesn't. They only share an EPR pair they created together, hence the coordination and ...
  • 2,413
8 votes
Accepted

How many bits do Alice and Bob needs to compare to make sure the channel is secure in BB84?

Your analysis of Eve's cheating doesn't seem quite right (although the final answer is correct). What you need to say is: Assume Alice prepares a particular state in one of the bases. You could assume ...
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7 votes
Accepted

Quantum entanglement for faster-than-light (FTL) network communication?

As this is one of the first examples in Nielsen & Chuang, I'll go ahead and type out their explanation here for anyone else that is interested in entanglement for faster than light communication. ...
  • 1,689
7 votes
Accepted

Protocol for entaglement swapping

For two parties to share an EPR pair means that each party has one qubit, and these two qubits together are in state $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ (or one of the other Bell pairs). ...
7 votes
Accepted

Quantum teleportation: second classical bit for removing entanglement?

Your initial calculations are correct. When Alice performs her first measurement and gets a 0 outcome then, as you say, Alice and Bob are left sharing a two-qubit state $$ |\Psi\rangle=\alpha|00\...
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7 votes

What is the practical interest of superdense coding?

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 ...
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7 votes
Accepted

How to prove that $\frac{| x_0 \rangle + | x_1 \rangle}{\sqrt{2}}$ hides one of $x_0$ or $x_1$?

We can bound the amount of information that can be retrieved from $|\psi\rangle$ using Holevo's bound. Alice and Bob Let us first reformulate the situation in the terms usually employed in the context ...
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6 votes
Accepted

Can superdense coding be made more efficient?

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 ...
  • 24.8k
6 votes
Accepted

Generalising the standard quantum teleportation protocol: what are the possible unitaries?

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 ...
  • 4,598
6 votes

Is it possible to have Quantum Teleportation Protocol without using classical channel?

This "protocol" is completely wrong. The author is simply preparing the state $|\psi\rangle$ on Bob's side and leaving it there untouched. Meanwhile, Alice prepares a maximally entangled ...
5 votes

Transmission of information over long distances

No. As far as we know, entangled states do not permit faster than light communication. You might be able to use them for things like doubling bandwidth (see superdense coding) or sending quantum ...
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5 votes
Accepted

How to transfer non maximally entangled state to maximally entangled?

The first question that we have to deal with is what is meant by "maximally entangled" in this context. There's no single straightforward notion. In particular, for three qubits, there are two ...
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5 votes

Advances in Quantum Channel Capacity

Let's recap a bit: In classical information theory, the analogous formula is the Shannon noisy channel coding theorem. It's charming, because it is basically just a very simple optimization of the ...
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5 votes
Accepted

Does the Lieb-Robinson bound constrain the speed of entanglement information transmission?

Entanglement does not transmit information, as follows from the No-communication theorem. Lieb-Robinson bound is a limit on speed at which perturbation propagates using short-range interactions, for ...
  • 3,124
5 votes

Quantum communications and "knowledge" of receiving a qubit

If you are talking about the idea that the quantum state is encoded on a physical system (perhaps an atom), and that system can be sent from Alice to Bob, then yes, you can detect the presence of the ...
  • 48.2k
4 votes

BB84 Protocol Alice Choice to Bob

That’s the public discussion stage: Alice and Bob can both announce which basis they chose for each round. If they happened to pick the same basis on a given round, they know that (in a perfect world) ...
  • 48.2k
4 votes
Accepted

Applying CNOT with local operations and two EPR pairs

Here is an idea of how could this be solved. It is based on teleportation. First, Alice teleports her qubit by means of one of the EPR pairs that she shares with Bob. In order to do that, she sends ...
4 votes
Accepted

Collective measurements: importance and realization

Collective measurements are normal measurements. You just need to be clear on the setting under which they are operating. I haven't delved deeply into the specific paper you mention (so it's always ...
  • 48.2k
4 votes
Accepted

Eavesdropping in superdense coding: why can't the third party infer anything about the message?

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 ...
  • 48.2k
4 votes

Does entanglement allow communication of user specified information or not?

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 ...
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4 votes

Why is measurement needed in teleportation?

measurements in every circuit can be postponed or never performed in a circuit while achieving the same functionality of the circuit That's correct. But if the circuit involves two parties, this ...
  • 24.8k
4 votes

What is the quantum state transmitted to Bob in BB84 protocol?

Application of Hadamard gates changes states $|0\rangle$ and $|1\rangle$ followingly: $\mathrm{H}|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ $\mathrm{H}|1\rangle = \frac{1}{\sqrt{2}}(|0\...
4 votes

If Alice and Bob share only classical communication resources, is shared entanglement always equivalent to shared randomness?

Entanglement + a classical channels allows you to build a quantum channel using teleportation. Thus, adding a quantum channel does not give you additional power.
4 votes

Is it possible to have Quantum Teleportation Protocol without using classical channel?

No. If no classical information is transmitted from Alice to Bob, then Bob's state is completely unaffected by anything that happens on Alice's side. This means, in particular, that no information ...
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4 votes
Accepted

Can joint measurement be achieved in two labs far apart?

TL;DR No. Non-local gates and measurements between causally disconnected observers violate the no-signalling theorem and are therefore impossible. No-signalling theorem The following protocol allows ...
  • 14.8k
3 votes
Accepted

Why is measurement needed in teleportation?

To answer this, contrast quantum teleportation with the swap gate. Ignoring everything in the middle, the effect of quantum teleportation is to get from state $|\phi00\rangle$ to $|00\phi\rangle$. ...
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3 votes

Eavesdropping in superdense coding: why can't the third party infer anything about the message?

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 $|\...
  • 19.6k
3 votes

Does entanglement allow communication of user specified information or not?

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 ...

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