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7 votes
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What is meant by a "single-letter" expression for the quantum channel capacity?

In a series of individual works, Lloyd, Shor, and then Devetak developed what is known as the "LSD Theorem," which gives a formula for the quantum capacity of a quantum channel. The result ...
Condo's user avatar
  • 2,048
6 votes
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What exactly is the relation between the Holevo quantity and the mutual information?

Right, they are quite similar. The Holevo bound is a bound on the amount of accessible information between your quantum system and your classical system. The I(X;B) object written in the HSW theorem ...
The Yomster's user avatar
5 votes
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Classical capacity of quantum channel - Holevo quantity vs accessible information of a channel

No they are not the same. Given some quantum channel $\mathcal{N}$ we can consider an encoding map $\mathcal{E}$ and a decoding map $\mathcal{D}$ such that $\mathcal{C}_1= \mathcal{D}\circ \mathcal{N} ...
Rammus's user avatar
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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 ...
Martin's user avatar
  • 349
5 votes

Quantum capacity for serial composition of quantum channels

TL;DR Quantum capacity of $\mathcal{N}_2\circ\mathcal{N}_1$ can be anywhere between zero and the minimum of the quantum capacities of $\mathcal{N}_1$ and $\mathcal{N}_2$. Background Quantum capacity ...
Adam Zalcman's user avatar
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5 votes
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Understanding classical vs. quantum channel capacities

These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let $\Phi:\text{L}(\mathcal{X}) \...
John Watrous's user avatar
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4 votes
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Bounding diamond norm distance using probability of error in transmission of classical information

Intuition The expression $\|\mathcal{A} - \mathcal{I}\|_\diamond$ quantifies how close the channel $\mathcal{A}$ is to the identity channel $\mathcal{I}$ which is the channel that preserves quantum ...
Adam Zalcman's user avatar
  • 22.9k
3 votes

Experimental Realization of Superactivation of Quantum Capacity

The paper Superactivation of Multipartite Unlockable Bound Entanglement, presented the first experimental realization of the following superactivation: Alice and Charlie have zero entanglement. Bob ...
user1271772 No more free time's user avatar
2 votes

Does proving $Q^{(1)}(\mathcal{N}\otimes\mathcal{N})=Q^{(1)}(\mathcal{N})+Q^{(1)}(\mathcal{N})$ imply additivity for arbitrary $n$?

I found out the paper Quantum Channel Capacities (by Graeme Smith) were the author states: "Some entropic function f(N ) is shown to be an achievable rate, and its regularization equal to the ...
Josu Etxezarreta Martinez's user avatar
2 votes
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Additivity of degradable and anti-degradable quantum capacities

I have been able to find the answer to this question, so I will post it myself for anyone that would be interested. The result is proven in Useful States and Entanglement Distillation by Leditzky, ...
Josu Etxezarreta Martinez's user avatar
2 votes
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Why does entanglement not increase the classical capacity of a channel?

In my understanding, the key part for entanglement to increase capacity is to have a suboptimal channel. Suppose the input of you channel can take value in the set $X$, and note $G(X)$ the graph where ...
Serwyn's user avatar
  • 116
2 votes

Why does entanglement not increase the classical capacity of a channel?

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 ...
Patrick Fuentes's user avatar
2 votes
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Prove that the coherent information of an antidegradable channel is equal to zero

Let the channel be $\mathcal{N}_{A\rightarrow B}$ and its complementary channel be $\mathcal{N}^c_{A\rightarrow E}$. By Property 13.5.1 of [1], we have that $$Q(\mathcal{N})\geq 0.$$ Consider the map $...
rnva's user avatar
  • 842
2 votes
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Classical capacity of the quantum bitflip channel

The classical capacity of the quantum bitflip channel is 1. The states $|+\rangle\langle+|$ and $|-\rangle\langle-|$ are invariant under bitflip channel. Let $\mathcal{N}$ be the bitflip map. $$ \...
FDGod's user avatar
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1 vote
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Is there a notion of approximate entanglement breaking (EB) channels?

This work (https://journals.aps.org/pra/abstract/10.1103/PhysRevA.97.012332) may be relevant to your question. The authors have shown that "approximate additivity" holds for "...
Seiseki Akibue's user avatar
1 vote
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Increasing quantum channel capacity

We define the rate to be $R = \frac{\log_k |\mathcal{M}(n)|}{n}$ where $\mathcal{M}(n)$ is the set of messages you can successfully transmit using $n$ i.i.d. copies of the channel and $k$ is some ...
rnva's user avatar
  • 842
1 vote

Twirling of quantum states: Maximally entangled states

Let $|\psi\rangle = \frac{1}{\sqrt{d}}\sum_{i=0}^{d-1} |ii\rangle_{AB}$ be a maximally entangled state on a bipartite system $AB$. Such states satisfy the transpose property $$ (X \otimes I) |\psi\...
Rammus's user avatar
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