19 votes
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How can classical bits be copied if qubits cannot be copied?

TL;DR: The ban on copying is not nearly as universal as it might seem. No-cloning theorem actually allows copying as long as it is limited to orthogonal states. Classical information is the type of ...
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11 votes
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Proof of an Holevo information inequality for a classical-classical-quantum channel

It appears that the statement is not true in general. Suppose $X = Y = \{0,1\}$, $\mathcal{H}$ is the Hilbert space corresponding to a single qubit, and $W$ is defined as \begin{align} W(0,0) & = |...
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  • 4,413
10 votes
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Is the set of all states with negative conditional Von Neumann entropy convex?

The conditional von Neumann entropy is a concave function: if $\rho$ and $\sigma$ are states of a pair of registers $(\mathsf{X},\mathsf{Y})$ and $\lambda\in[0,1]$ is a real number, then $$ \mathrm{H}(...
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  • 4,413
9 votes

Is the set of all states with negative conditional Von Neumann entropy convex?

Geometric characterization (as any other characterization) of subsets of the quantum state space in relation with their locality and entanglement properties becomes very complicated as the number of ...
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8 votes
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Partial trace over a product of matrices - one factor is in tensor product form

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the ...
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  • 4,413
8 votes
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Closeness of purifications of states

No dimension-independent bound is possible. Consider states $\rho_A$ and $\sigma_A$ that are close in $p$-norm (for $p>1$) but have relatively low fidelity. Specifically, assume $$ \|\rho_A - \...
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  • 4,413
7 votes
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Degradable channels and their quantum capacity

A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$. The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is ...
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  • 4,413
7 votes

Partial trace over a product of matrices - one factor is in tensor product form

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the ...
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7 votes
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What does it mean to take the Choi-Jamiolkowski of a quantum channel?

Let me quote my answer from over at physics.SE: The intuition Let us consider a channel $\mathcal E$, which we want to apply to a state $\rho$. (This could equally well be part of a larger system.) ...
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7 votes
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What is the Von Neumann entropy of $\rho = \sum_ip_i|i\rangle\langle i| \otimes \rho_i$?

Operator $\rho$ is not a tensor product, it's a sum of tensor products $$ p_1|1\rangle\langle 1| \otimes \rho_1 + p_2|2\rangle\langle 2| \otimes \rho_2 + \dots + p_d|d\rangle\langle d| \otimes \rho_d. ...
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  • 5,423
7 votes
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Stinespring dilation: Size of environment

No. The minimal size of the environment is just the rank of the Choi matrix of $\mathcal E$, call it $J(\mathcal E)$. Since $J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n}$ and $\text{...
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7 votes
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Does the no-hiding theorem suggest that quantum information is never destroyed?

It is true that unitary evolution cannot destroy information. This is the content of the no-cloning theorem and its time reversal - the no-deleting theorem. The no-hiding theorem says something ...
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7 votes
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Schmidt decomposition for tripartite system $ABC$ with vanishing mutual information between $A$ and $C$

TL;DR: The key observation is that Schmidt basis on a subsystem consists of eigenvectors of the reduced state of that subsystem. Consequently, if the reduced state is a product state then its Schmidt ...
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7 votes
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Existence of a perturbed channel that achieves a perturbed output state

Yes, the channel $\tilde{N}$ necessarily exists. Notice first that the state $\rho_B$ is the completely mixed state $\mathbb{1}/d$. So, in order for $\tilde{\rho}_{A'B}$ to be contained in $S$, three ...
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  • 4,413
6 votes

Violation of the Quantum Hamming bound

You may be interested in the answers to this question. One example of a degenerate code beating the quantum Hamming bound is here. I also have a numerical example of a small violation in my own work, ...
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6 votes
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Accessible information of system vs system, apparatus and environment

For density matrices $\rho_A$ and $\rho_B$ having eigenvalues $\lambda^{\left(A\right)}$ and $\lambda^{\left(B\right)}$, \begin{align}S\left(\rho_A\otimes\rho_B\right) &= -\rho_A\otimes\rho_B\ln\...
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  • 3,397
6 votes
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How many classical bits are needed to represent a qubit?

There are two types of information in physics: Classical information Quantum information Physics doesn't answer the question "What is (classical or quantum) information?". This is philosophic ...
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  • 3,004
6 votes
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If quantum computing always return random measurement (or uncertain measurement), why do we still need it?

Short answer: Assuming you are measuring in the computational basis (Z basis), $\{|0\rangle, |1\rangle \}$, there is no randomness upon measurement in the following quantum circuit (you will always ...
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  • 12.3k
6 votes
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Can one quantify entanglement between different parts of a system?

Of course you can. Take any entanglement measure that can be applied to a system whose overall description is a density matrix, and you can apply that to the density matrix describing your subsystem. (...
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  • 46.1k
6 votes

How to describe a known quantum state using classical information?

Generally speaking, in order to describe elements of a set $A$ using classical information we need two ingredients: a non-empty finite alphabet $\Sigma$ and an encoding $E: A\to\Sigma^\omega$ which ...
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6 votes
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Can all mixed states be written as a convex combination $\rho=\sum_j p_j |\psi_j\rangle\langle \psi_j|$?

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} ...
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  • 876
5 votes
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Quantum circuit for computing fidelity

Prohibited device Such a circuit $C$ would enable faster-than-light communication and therefore does not exist. Suppose Alice and Bob share a Bell pair $|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |...
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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}) \...
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  • 4,413
5 votes

Relating min-entropy with conditional entropy

The conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ can be defined for an arbitrary state $\rho$ of a pair of registers $(A,B)$ as $$ - \inf_{\sigma} \,\text{D}_{\text{max}}(\rho \| \...
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  • 4,413
5 votes
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Does computing the quantum mutual information $I(\rho^{AB})$ require full tomographic information of $\rho^{AB}$?

The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation: $$I(...
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5 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 ...
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5 votes
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Can classical linear algebra solvers implement quantum algorithms with similar speed-ups?

The complexity that you are glossing over is that in the general case you need to store $2^n$ complex amplitudes to even represent an $n$ qubit system classically. Therefore, for a quantum computer of ...
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5 votes
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Does the quantum Jensen-Shannon divergence appear in any quantum algorithms or texts on quantum computing?

That quantity appears to be identical to Holevo information, which turns out to be the upper bound on how much classical information you can transmit using a quantum channel [1]. More generally the ...
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  • 4,507
5 votes
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Fidelity of extensions of states

Let's start with the second question. There is nothing special about an extension $\sigma_{AR}^{\ast}$ that allows it to be optimal for the right-hand side of (1); any extension $\sigma_{AR}$ of $\...
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  • 4,413
5 votes
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Prove the additivity of the Renyi entropy: $H_{\beta}(p \times r) = H_{\beta}(p) + H_{\beta}(r)$

This only holds if the two distributions are independent. In this case $$ \begin{aligned} H_{\beta}(p \times q) &= \frac{1}{1-\beta} \log\left( \sum_{i,j}(p(i) q(j))^{\beta} \right) \\ &= \...
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