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14 votes

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

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 ...
6 votes
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Is the trace of a positive map always positive?

Consider the qubit map $\Phi(\rho):=\sigma_Y\rho^T\sigma_Y$, that is, $$ \Phi\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}=\begin{pmatrix}\rho_{22}&-\rho_{12}\\-\...
Frederik vom Ende's user avatar
6 votes

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

Prove that if Kraus operators of $\Phi$ form an ONB then $\Phi$ is the replacement map

This solution uses the spectral theorem and some elementary linear algebra computations. If you let $P = \sum_{i = 1}^{d'} \lambda_i \vert{\psi_i}\rangle\langle{\psi_i}\vert$ be the spectral ...
user2533488's user avatar
5 votes
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Can any channel be represented as $A\rho A^\dagger$ for some $A$?

No, this is not possible in general. To see it, consider for example what happens taking the trace of that expression. You'd get: $$\operatorname{tr}(A^\dagger A\rho)=\operatorname{tr}\left(\sum_j K_j^...
glS's user avatar
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5 votes

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
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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
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 are kraus operators of a qubit interacting a thermal environment?

One way to model interaction with a thermal environment (of some temperature $T$) is through so-called thermal operations. Given some system's Hamiltonian $H_S$ they're all channels $\Phi$ of the form ...
Frederik vom Ende's user avatar
4 votes

Can Kraus operators change a mixed state into a pure state?

To add to the already great answers let me point out that if the initial mixed state $\rho$ has full rank, then there is only one type of channel which can send that state to a pure state $|\phi\...
Frederik vom Ende's user avatar
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What can we say about $\sum_i K_i K_i^\dagger$ for non-unital CPTP maps?

Let $d$ be the dimension of the system. Since $\sum_i K_iK_i^\dagger = \Phi(\mathbb 1) = d \Phi(\mathbb 1/d)$, it can be $d$ times any quantum state (by, e.g., setting $\Phi$ as a replacement channel)....
Senrui Chen's user avatar
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How many free real parameters in a general CPTP map?

An easier way of counting is to consider the Choi state: It is a hermitian $d_Ad_B\times d_Ad_B$ matrix, and thus has $(d_Ad_B)^2$ real parameters. The condition that the reduced state of $A$ is the ...
Norbert Schuch's user avatar
4 votes
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Prove that if Kraus operators of $\Phi$ form an ONB then $\Phi$ is the replacement map

Distraction The matrix $P$ is a distraction, because the $dd'$ matrices $A_k$ are an orthonormal basis with respect to the inner product $\langle A,B\rangle_P:=\mathrm{tr}(PA^\dagger B)$ if and only ...
Adam Zalcman's user avatar
4 votes

Why do we need/have the operator sum representation (Kraus representation)?

To add to Daniele's answer, $E_k$ is an operator—and not a scalar—because the notation in Nielsen & Chuang is sloppy. What is meant is that $E_k=\iota_k^\dagger U\iota_0$ where $\iota_k:\mathbb C^...
Frederik vom Ende's user avatar
3 votes
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Affine transformation of the Bloch sphere to Kraus representation of qubit channels

TL;DR: The given form of the channel is essentially an obscured way of writing down the channel's Pauli transfer matrix. One set of Kraus operators coincides (up to vectorization) with the ...
Adam Zalcman's user avatar
2 votes

Why do we need/have the operator sum representation (Kraus representation)?

The formalism is useful to describe a noisy transformation $\mathcal{E}$ only in the domain of a system of interest $\mathcal{H}$. According to the book, you can select an orthogonal bases $\{e_i\}_i$ ...
Daniele Cuomo's user avatar
2 votes

Why is the Choi matrix different from the analytic form for a depolarizing channel?

Your definition of the depolarizing channel is slightly off, it should read $$ \mathcal E(\rho)=(1-p)\rho+p\,\boxed{{\rm tr}(\rho)}\frac I2\,. $$ This extra term ${\rm tr}(\rho)$ is necessary for $\...
Frederik vom Ende's user avatar
2 votes
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How to get the Kraus decomposition of the amplitude damping channel from its Choi?

Decomposing $C_{N_\gamma}$ amounts to finding all non-zero eigenvalues as well as corresponding eigenvectors: Assume for now that we already found all eigenvalues $\lambda_j>0$ as well as ...
Frederik vom Ende's user avatar
2 votes

Find the Kraus operators for the amplitude damping channel from its isometric representation

The question is essentially about how to relate the unitary/isometric representation of the dephasing channel with its representation in terms of Kraus operators. In general, an isometric ...
glS's user avatar
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2 votes
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Proof that two sets of quantum maps are equivalent only when they are related by a unitary transformation

Let's take as a given that $A^\dagger A=I$, and that $A$ is square (that just means being careful about indices in your derivation, and perhaps introducing some padding of 0 operators). What is $AA^\...
DaftWullie's user avatar
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2 votes

What can we say about $\sum_i K_i K_i^\dagger$ for non-unital CPTP maps?

One can show concretely that a $\tfrac1d\Phi(I)$ can be any density matrix $\sigma$. To this end, simply choose $$ \Phi(\rho) = \mathrm{tr}(\rho)\,\sigma\ . $$ It can be straightforwardly checked it ...
Norbert Schuch's user avatar
2 votes

Why do we need/have the operator sum representation (Kraus representation)?

To add to Daniele's and Frederik's answers: the operator sum representation is even more useful than the Nielson and Chuang derivation might suggest - though, the derivation is very helpful from a ...
qubitzer's user avatar
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2 votes

How does measuring a density matrix give Kraus operators?

TLDR If you measure a qubit in the computational basis without getting the result of the measurement the corresponding channel can be described by the Kraus operators $K_0 = |0\rangle \langle 0|$ and $...
qubitzer's user avatar
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1 vote
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How to get the Kraus operator $M_0=\sqrt{1-p}\, I$ for the depolarizing channel, from its isometric representation?

The unitary representation specifies how the isometry $U$ acts on pure states. The channel $\Phi$ is related to $U$ by $\Phi(\rho)=\operatorname{tr}_E[U\rho U^\dagger]$. Notice that on pure states ...
glS's user avatar
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1 vote

Get matrix for an X gate for a given fidelity p

The usual way to represent this isn't via a matrix, but via the Kraus representation. That is, a quantum operation can be written as: $$\rho\mapsto\sum_iK_i\rho K_i^\dagger$$ with $\sum\limits_iK_i^\...
Tristan Nemoz's user avatar
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1 vote

Can a unital channel not be mixed unitary?

Unital channel which is not mixed unitary Consider the following two-qubit quantum channel $$ \Xi(X)=A_0XA_0^\dagger+A_1XA_1^\dagger\tag1 $$ where $$ A_0=\begin{bmatrix} 1&&&\\ &&\...
Adam Zalcman's user avatar
1 vote
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Interconversion between different representations of quantum channels

The only question which I believe has not been addressed by the comments yet is your question 3: how is the Stinespring representation related to the representation matrix $K(\Phi)$ of $\Phi$? To ...
Frederik vom Ende's user avatar
1 vote
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What is the relation between the Choi matrix and the Liouville space (superoperator) representations of a channel?

I think there's some confusion here, so let me try to clarify some basic things: Given any quantum channel $\Phi$, you define its Choi representation as the operator $J(\Phi)=(\Phi\otimes \...
glS's user avatar
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