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

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

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

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 Bob to send one classical bit $b$ to Alice faster than light. Alice prepares two sets of qubits $x_i$ and $y_i$ in the $|0\rangle$ state. Bob prepares one set of ...


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

First things first. The quantum teleportation protocol requires classical bits to be sent from Alice to Bob. Without these bits, Bob cannot apply the proper corrections to actually obtain the desired "target" state $|\psi\rangle$, this is apparent by understanding what's going on in the teleportation protocol. Additionally, the requirement of a ...


3

It's the dimension of the Hilbert spaces. In DV-QKD you have a finite dimensional Hilbert space (like a qubit). Thus your measurement outcomes come from a finite set. On the other hand a CV-QKD protocol uses infinite dimensional systems and therefore you can have a continuum of measurement outcomes. If there's no specification as to whether a QKD scheme is ...


2

Bob can't know in any sense that Alice has sent a message, until and unless Bob receives a classical message from Alice confirming that she has sent a qubit. Even for the sake of argument, let's suppose he somehow knew that Alice has sent a qubit, then arises two scenarios - one, Bob knew it instantly and second, Bob knows only after the time it takes for a ...


2

There are lots of different situations one can talk about from a cryptography perspective. But here's one that has a huge practical relevance: There are physical realisations of quantum computers which, individually, are limited in the number of qubits they can use. For example, ion traps. For the sake of argument, assume you can hold 10 qubits in each ion ...


2

There are two different complexity considerations that you might be interested in in this scenario. The one that most people think about is the communication complexity, i.e. how many qubits Alice and Bob have to send to the referee. This is relatively simple to calculate - it's just $mk$ where $k$ is the number of times you have to repeat the controlled-...


2

Not an exact answer to your question, but a possible way to get some upper bounds. Look into MacKay2003 (~page 29). He uses capacity of classical channels to get upper bounds for quantum channels. For classical channels, "capacity with allowable error rate" is known; so you can trace the process in the paper to get similar results for what you're ...


1

Yes, Alice and Bob can do better than option 1. However, Alice needs to send 2 classical bits to Bob. This is more than in option 2, but is still independent of $n$. The idea is that instead of performing a measurement Alice applies a unitary to rotate the measured subspace onto an auxiliary qubit and then teleports the auxiliary qubit to Bob who performs ...


1

Mathematically, if Alice measures her part of an entangled bipartite state (or does whatever she likes with her part), the local state of the Bob's part (described by density matrix) does not change. This means that no information is transferred from Alice to Bob by Alice manipulating her part of an entangled state, at all.


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