# Tag Info

## Hot answers tagged communication

6

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 $$|\Psi\rangle=\alpha|00\rangle+\beta|11\rangle$$ (you can safely ignore the measured qubit). The problem is the statement She can tell Bob: "Your half of the EPR is now the qubit I ...

5

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 states, but that will all happen at the speed of light (or slower). It is true that entangled states do seem to know something about the constituent elements faster ...

5

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). The entanglement swapping protocol is described, for example, in Wikipedia. It is the same as traditional (I almost wrote "classical" but caught myself in time) ...

4

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 possible they make marginally different assumptions), but I expect it goes like this: You are looking at using many copies of the same channel. Encoding will, ...

4

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 process will introduce quantum operations between the two parties. It will require a quantum communication channel, so that the qubits can be shuttled back and ...

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

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

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 $|00\phi\rangle$. This can obviously be accomplished with a simple application of the swap gate. So, why do we care about quantum teleportation? You'd never run the quantum ...

3

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 the classical information she obtains by measuring her EPR halve and her qubit. Bob uses the clasical information received in order to reconstruct the qubit in ...

3

There is a set of two packages in Mathematica called "Quantum Notation" and "Quantum Computing" for Wolfram Mathematica Environment, and here you can very well mimic all three considerations you are concerned with and much more in the usual Dirac Notation and Quantum Circuit Formalism. The link to the packages are as follows: http://homepage.cem.itesm.mx/...

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Reutter and Vicary Features of teleportation, dense coding and secure key distribution are mimicked even without having an honest quantum internet. The main idea is the groudit. That is a special type of groupoid where there certain bijections as sets given as extra data. That is for a given natural number $d$, think about $d$ finite groups all of ...

2

Absolutely! BB84 can be implemented over LiFi. There are however certain constraints and inefficiencies involved, which will have to be taken into account for implementation - possibly leading to the usage of BB84 variants like Decoy State, with higher efficiencies than pure BB84. A four non-orthogonal polarization basis encoded single photon would be very ...

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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 ...

1

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 ...

1

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

I wanted to expand on DaftWullie's accepted answer in case it is helpful to others. DaftWullie's is correct; mine is only a continuation of the same idea. Consider the one-qubit state $\psi = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. We know that applying the Hadamard matrix $H = \frac{1}{\sqrt{2}}\begin{bmatrix}1& 1 \\ 1 & -1\end{bmatrix}$ will ...

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