12

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 are called quantum communications. These problems/situations sometimes disambiguate by calling their qubits "flying qubits" which are almost always photons. ...


11

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 paper. In this paper, he proposed five criteria, which are widely considered to be the five (sufficient and necessary) criteria that any physical quantum computer ...


10

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 after, what Alice sends to Bob when communication takes place are classical bits, she sends the measured bits of the 2 qubits she has, so the only communication we ...


8

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 that's $|0\rangle$, but you can make the argument more generally. With 50% probability, Eve measures in the same basis that Alice prepared in (the 0/1 basis ...


7

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. The following is an abridged version of Nielsen & Chuang section 1.3.7 entitled 'Example: quantum teleportation' Quantum teleportation is a technique ...


7

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 ...


7

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) ...


7

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 of Holevo's bound. Suppose Alice chooses two $n$-bit strings $x_0$ and $x_1$ independently and uniformly at random. Let $X$ represent the random variable ...


6

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 ...


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 ...


6

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 state, and applies a CNOT and a Hadamard to it, which maps it into the state $|00\rangle$. It is not possible to teleport without a channel.


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 ...


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

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 inequivalent classes of entangled state that cannot be interconverted by SLOCC (stochastic local operations and classical communication). Each has a maximally ...


5

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 mutual information. The quantum channel capacity is that it is given by $$ \lim\limits_{n\to\infty} \frac{1}{n}Q(T^{\otimes n}) $$ where $T$ is the quantum ...


5

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 atom without measuring the state of the atom. To make the point, I'm going to go a bit crazy. I'm not claiming this is exactly a physical scenario.... Imagine ...


4

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) their answers were the same, so they can translate them into a 0/1 value that nobody else knows. That translation is arbitrary, and they’ve probably agreed it ...


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

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 ...


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 ...


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 ...


4

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 example in spin lattice. I doubt it means something for protocols such as BB84; you can transmit quantum information by sending polarized photons, and photons ...


4

Application of Hadamard gates changes states $|0\rangle$ and $|1\rangle$ followingly: $\mathrm{H}|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ $\mathrm{H}|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ Identity operator does not change the state in any way, i.e. $\mathrm{I}|0\rangle = |0\rangle$ and $\mathrm{I}|1\rangle = |1\rangle$. Hence ...


4

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

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 whatsoever can be sent using such protocol. More precisely, if $\rho$ is the initial bipartite (presumably entangled) state, then the only thing Bob has access to ...


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

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 ...


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

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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