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4 votes

Does $N(U\rho U^\dagger)=U' N(\rho)U'^\dagger$ for unitaries $U,U'$ and a channel $N$ imply $UK_i=K_i U'$?

As Markus Heinrich points out, the answer is "no". On the other hand, for an abelian symmetry group, you can find a set of Kraus operators such that each Kraus operator individually ...
Norbert Schuch's user avatar
4 votes
Accepted

Does $N(U\rho U^\dagger)=U' N(\rho)U'^\dagger$ for unitaries $U,U'$ and a channel $N$ imply $UK_i=K_i U'$?

No. Take a depolarizing channel, which can be written as $$ \mathcal{D}_p(X) = p\, \mathrm{tr}(X) \frac{I}{d} + (1-p) X\, $$ Note that it does commute with any unitary $U$, $\mathcal{D}_p(U X U^\...
Markus Heinrich's user avatar
1 vote
Accepted

Are peripheral eigenvalues of a completely positive map always semisimple?

Consider $$ K:=\begin{pmatrix}1&1\\0&1\end{pmatrix} $$ as well as $\Phi:=K(\cdot)K^\dagger$. This map is completely positive (because $\Phi$ is in Kraus form) and even strictly positive ...
Frederik vom Ende's user avatar
1 vote
Accepted

Are Stinespring unitaries that give rise to the same channel locally unitarily equivalent?

As it turns out the statement in question is wrong. For a counterexample consider the full-rank environment state $$ \omega=\begin{pmatrix} \frac12&0&0\\0&\frac14&0\\0&0&\...
Frederik vom Ende's user avatar
0 votes
Accepted

Given $\Psi$ completely positive when do there exist $K_1,K_2$ such that $K_2\Psi(K_1^\dagger(\cdot)K_1)K_2^\dagger$ is also trace preserving?

For a counterexample consider the qubit map $$ \Psi\begin{pmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\end{pmatrix}:=\begin{pmatrix}x_{11}&0\\0&0\end{pmatrix}. $$ This map is completely ...
Frederik vom Ende'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
  • 507
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
  • 507
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
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

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