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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 ...
9 votes

Can a CPTP map increase the purity of a state?

Yes, some quantum channels can increase purity. For example the preparation channel $$ T(X) = \mathrm{Tr}[X] |\psi\rangle \langle \psi| $$ that can be thought of as throwing away your system and ...
Rammus's user avatar
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Why is the coefficient-squared the probability, and not just the coefficient itself?

Scott Aaronson describes quantum mechanics as "statistics but with the L2-norm". States are L2-norm unit vectors (sum of squared amplitudes is 1) instead of L1-norm unit vectors (sum of ...
Craig Gidney's user avatar
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How can the depolarizing channel be a quantum operation?

The correct linear form of the depolarizing channel is $$ \varepsilon(\rho) = (1-p)\rho + p\frac{I}{2}{\rm Tr}(\rho). $$ For density matrices ${\rm Tr}(\rho)=1$, so you can usually see the form ...
Danylo Y's user avatar
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What channels preserve the purity of all pure inputs?

TL;DR: The unitary and reset channels are the only ones that return pure output for every pure input. That's because under Stinespring dilation the requirement that $\Phi$ return pure output for every ...
Adam Zalcman's user avatar
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What are "completely positive" and "CPTP" quantum maps?

[A] States lie in Hilbert space $\mathcal{H_S}$. $|\psi\rangle \in \mathcal{H_S}\,.$ Operators, density operators lie in the bounded operator space of $\mathcal{H}_S$. $\rho \in \mathcal{B}(\...
FDGod's user avatar
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What can we say about the eigendecomposition of quantum channels?

As you observe correctly, $\mathbb N$ is a linear map. Thus, the same holds as for any eigendecomposition of linear maps. In particular, there need not be a complete basis of eigenvectors (there can ...
Norbert Schuch's user avatar
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 ...
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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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Does a quantum channel always preserve the identity matrix?

No, such maps are referred to as unital maps. A counterexample is the replacement map $$ \mathcal{E}(X) = \mathrm{tr}(X)\sigma $$ defined for some density matrix $\sigma$.
Rammus's user avatar
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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. ...
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Existence of Hamiltonians such that the time evolution unitary becomes identity

I don't believe that this is always possible. For instance, what if my set of $\{H_i\}$s comprise a single term that I can construct to be arbitrarily awkward? The key feature will be gaps between ...
DaftWullie's user avatar
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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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Qiskit reverse_bits is not equivalent to swapping qubits

The issue is that the qubit values are being mixed together throughout the circuit, so swapping at the end is not enough. However, if you also swap at the beginning of the circuit, the qubit values ...
Nick Mertes's user avatar
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

Resources for understanding non-unitary channels and operators

Lecture Notes on the Theory of Open Quantum Systems by Lidar The Theory of Quantum Information by Watrous Principles of Quantum Communication Theory: A Modern Approach by Khatri & Wilde
FDGod'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

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
4 votes

Interesting properties of maps whose natural representation is unitary?

It may or may not come at a surprise that the natural representation of channels can only be unitary if the channel itself is unitary. More precisely, we will prove the following statement: Theorem. ...
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
4 votes

What are "completely positive" and "CPTP" quantum maps?

Definitions Let $\mathcal{H}$ be a complex Hilbert space. It turns out that the set $L(\mathcal{H})$ of all linear operators on $\mathcal{H}$ is also a Hilbert space. Let $I_\mathcal{H}$ denote the ...
Adam Zalcman's user avatar
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How do you work out the matrix for controlled-U operations?

$\left|0\right>$ and $\left|1\right>$ don't mean $\begin{pmatrix}0 \\ 0\end{pmatrix}$ and $\begin{pmatrix}1 \\ 1\end{pmatrix}$. In bra-ket notation we usually fix some basis and denote by $\left|...
Vladimir Lysikov's user avatar
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Is the "unitary twirling operation" physically realizable?

Question 1: I guess it depends what your understanding of "physical" is. In my understanding, everything you can do in the lab is physical. Thus, twirling is perfectly physical. Note that ...
Markus Heinrich's user avatar
4 votes

Evolution of a state vector: Why is the action of $N$ equivalent to the action of $UNU^{†}$?

Maybe it's easier to see when it's presented like this: $UN\left|\psi\right> = UNU^\dagger U\left|\psi\right>$ means that $UNU^\dagger$ maps $U\cdot \left|\psi\right>$ to $U \cdot N\left|\psi\...
Vladimir Lysikov's user avatar
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Evolution of a state vector: Why is the action of $N$ equivalent to the action of $UNU^{†}$?

I think it probably helps to understand what Gottesman is trying to do with the operator $N$ (later in the paper). He wants to start with some state $|\psi\rangle$, but instead of directly describing ...
DaftWullie's user avatar
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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
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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

Does a quantum channel always preserve the identity matrix?

No. When this is the case the channel is said to be unital. A necessary and sufficient condition for a map $\mathcal E$ being unital is its adjoint $\mathcal E^\dagger$ being trace-preserving. For ...
glS's user avatar
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4 votes

What can we say about the eigendecomposition of quantum channels?

At the risk of saying something trivial I'd like to add to Norbert's great answer and point out that normal channels are necessarily unital, i.e. to have an orthonormal basis of eigenvectors it is ...
Frederik vom Ende's user avatar

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