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Counterexamples in quantum information theory

Quantum Channels Quantum channels: general properties Not every positive map is completely positive. One may argue that this is the mother of all counterexamples in quantum information: the ...
7 votes

How can quantum error correction correct small rotations/continuous errors?

The trick is to rewrite your continuous rotations as a perturbative sum. For example, consider applying $R_Z(\theta)$ to all data qubits of an $n$-qubit code. You can rewrite: $$R_Z(\theta) = I \cos \...
Craig Gidney's user avatar
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Counterexamples in quantum information theory

Quantum Computing / Quantum Complexity Theory Requirements for exponential speedup Clifford circuits can (1) create superposition such as with Hadamard gates, (2) create entanglement such as with ...
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Equal partial traces

No, assume $\rho_{AB}$ is pure, so that $\rho_{AB} = |u\rangle\langle u|_{{AB}}$. Since it's pure $\rho_{ABC}$ must have the form $\rho_{ABC} = \rho_{AB} \otimes \rho_{C}$. It follows that $\rho_{AC} =...
Danylo Y's user avatar
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Counterexamples in quantum information theory

General quantum information Entropies The relative entropy of entanglement is not additive, see Section V.B of this paper (arXiv) for a counterexample The minimal output entropy is not additive. ...
5 votes

Can we combine the square roots inside the definition of the fidelity?

This result is indeed correct and gets rediscovered from time to time. We rediscovered it last year and got it published at Phys Rev A 107, 012427 (2023) (free version at arXiv:2211.02623). But it's ...
Jonathan Jones's user avatar
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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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Counterexamples in quantum information theory

Quantum states Quantum states: general properties The purifications of two $\varepsilon$-close states need not be $\varepsilon$-close. The fidelity depends on more than just the difference of states. ...
5 votes

Counterexamples in quantum information theory

Quantum error correction A quantum error correcting code that corrects every single-qubit X and Z error need not correct every single-qubit Y error. Not any 3-colorable lattice can be used to create ...
4 votes

Counterexamples in quantum information theory

Quantum thermodynamics The set of thermal operations is not (topologically) closed. In the qubit case, the set of channels which lie arbitrarily close to the thermal operations is characterized in ...
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What is the meaning of $\sum_i K_iK_i^\dagger$ for a quantum channel with Kraus operators $K_i$?

Yes, the sum $\sum_iK_iK_i^\dagger$ is (a scalar multiple of) the quantum state output by the channel on the maximally mixed input \begin{align} N\left(\frac{I}{d}\right)=\frac{1}{d}\sum_iK_iK_i^\...
Adam Zalcman's user avatar
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If $\rho_{AB}$ is a separable then the partial transpose w.r.t to A is PSD

The main result that you need to complete all the steps that you mention is that if $\rho$ is a density matrix, then $\rho^T$ is also a density matrix. So, what are the key properties of a density ...
DaftWullie's user avatar
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Can post-measurement states have entropy larger than the original state?

The stronger inequality holds if $\rho$ is a pure state. You can see it from the fact that $\sqrt M \rho \sqrt M$ has unit rank if $\rho$ does. Thus in all such cases you have $S(\rho)=S(\rho_i)=0$ (...
glS's user avatar
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Does independence from the input state imply a tensor product structure for the unitary?

You can frame this as a question about complementary channels to characterise if and when the statement applies. For starters, let me recall a few basic facts: Any channel $\Phi$ can be represented ...
glS's user avatar
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Does independence from the input state imply a tensor product structure for the unitary?

If you apply a controlled-$V$, controlled off system $B$ and targeting system $A$, then because system $B$ is in $|0\rangle$, it will do absolutely nothing. Thus, the output after tracing will be ...
DaftWullie's user avatar
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Equal partial traces

As noted in the comment, the state part of the question is precisely the question of symmetric extensions; see for instance the thesis https://arxiv.org/abs/1103.0766 and references therein. Apart ...
helloworld's user avatar
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Erasure errors in quantum error correction

You should think of error correction as the process of measuring syndromes (i.e. determining the $\pm 1$ values of stabilisers, which is the case I'll exclusively focus on) and then following a lookup ...
DaftWullie's user avatar
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What is the difference between classical-quantum and completely classical states?

Your expressions give a pretty clear distinction: in the classical-quantum state, the eigenbases $\left|y_x\right>$ can be different for different states $x$ of the classical register $A$, and in ...
Vladimir Lysikov's user avatar
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How to compute the QFI of a thermal state?

QFI must always be computed with respect to a parameter "$\theta$". Perhaps it is the temperature that you want to use here, or $\beta$? Regardless, we can put the state into your desired ...
Quantum Mechanic's user avatar
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Deriving the choi matrix definition of the quantum depolarizing channel

For $\mathcal E(X)=p\mathrm{tr}(X)\,\frac{\mathbb I}{2}+(1-p)X$, then $$\begin{align} \sigma &= (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle \Omega|)\\ &= \sum_{ij} \mathcal E(|i\...
HerrWarum's user avatar
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How to prove there's no quantum channel that clones all classical states?

Consider a classical state $$ \rho = \sum_x p(x) |x\rangle \langle x|\,. $$ Now take a quantum channel $\Phi(M) = \sum_x K_x M K_x^\dagger$ defined by the Kraus operators $\{K_x\}_x$ where $K_x = |xx\...
Rammus's user avatar
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Are all extremal points of the feasible set of an arbitrary affine equation pure states?

A simple counterexample but perhaps I'm misinterpreting "one affine equation". Take the map $\Lambda$ to be the identity map and let $Y$ be any mixed state. Then the set of states satisfying ...
Rammus's user avatar
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3 votes

Does measurement in different bases allow for FTL communication?

If Alice measures in the $|+ \rangle$, $| - \rangle$ basis, then Bob's qubit will indeed be in either the $|+\rangle$ or $|-\rangle$ state depending on the result of Alice's measurement. But your ...
Nick Mertes's user avatar
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How can quantum error correction correct small rotations/continuous errors?

When you encode in an error correcting code, you select a subspace that your encoded qubit sits in. You can identify this with some projector $P$. For instance, if you know your logical 0 and logical ...
DaftWullie's user avatar
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2 votes

Can we combine the square roots inside the definition of the fidelity?

The reasoning of OP seems correct to me. I think it is best understood in terms of eigenvalues. Quantum fidelity, typically defined as $$ F(\rho,\sigma) = \text{Tr}\left(\sqrt{\sqrt{\rho}\sigma\sqrt{\...
Adrian Müller's user avatar
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How to justify the conclusion $|E_{sq}(\rho)-E_{sq}(\sigma)|\le f(\epsilon)$, when proving the continuity of the squashed entanglement?

Regarding the dimensions, this is not a problem because we can purify $\rho$ and $\sigma$ using the same system and then we anyway consider applying the same map to both systems and hence the ...
Rammus's user avatar
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Quantum relative entropy between pre- and post-measurement states

Firstly notice that the divergences are $$ D(\rho\|\rho_{\mathrm{rank-one}}) = - S(\rho) + S(\rho_{\mathrm{rank-one}}) $$ and $$ D(\rho\|\rho_{\mathrm{coarse}}) = - S(\rho) + S(\rho_{\mathrm{coarse}}) ...
Rammus's user avatar
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Is Klein's inequality due to Klein?

As far as I can tell Klein never published the proof but the proof is rather attributed to him via word of mouth. In the paper Mean Entropy of States in Quantum‐Statistical Mechanics (free version ...
Rammus's user avatar
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2 votes

How to discriminate between $N$ states drawn from one of two ensembles?

Unless I'm mistaken, your special case is also the solution to the generalization, up to the fact that you have to redefine $\rho$ and $\sigma$. Forget about $Q$ for now. If Alice gives Bob one copy ...
Tristan Nemoz's user avatar
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quantum generalisation of random variables

If I understand your question, the way a classical random variable $X$ with support $\left[2^n\right]=\left\{0,\cdots,2^n-1\right\}$ is represented in quantum information is via a diagonal density ...
Tristan Nemoz's user avatar
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