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5 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
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

Definition of a quantum gate

Sometimes people mean unitary-only. Sometimes they mean generically anything you might do to the qubits (like a measurement gate or a reset gate or a dynamical decoupling gate or a leakage removal ...
Craig Gidney's user avatar
  • 38.8k
3 votes
Accepted

Do all Hermiticity-preserving maps generate completely positive maps?

First a basic observation: if all Hermitian preserving $\mathcal L$ gave rise to completely positive dynamics $e^{t\mathcal L}$ for all $t\geq 0$, then so would $-\mathcal L$ (still Hermitian ...
Frederik vom Ende's user avatar
3 votes

Definition of a quantum gate

I wouldn't say it's coupled with the measurement gate. I think they are inherently different: A quantum gate is equivalent to a Unitary matrix. Quantum gates create together a quantum circuit (which, ...
Nati Erez'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
  • 517
2 votes
Accepted

For how many different times do I have to know that $e^{tL}$ is a quantum channel to conclude that $L$ is of Lindblad form?

Consider the "pancake channel" $$ \mathcal P:\rho \mapsto \mathrm{tr}\rho\,I + \sum_{\alpha=x,y,z} \lambda_\alpha\mathrm{tr}(\rho\sigma_\alpha)\sigma_\alpha\ , $$ where we choose $\...
Norbert Schuch's user avatar
2 votes

What is the complementary map of a serial concatenation of quantum channels?

The formula has multiple problems: first, complementary channels are of course not unique and, even worse, the output dimension is not fixed; so to make things rigorous we would have to impose some ...
Frederik vom Ende's user avatar
2 votes

What is the domain of the dual map of a quantum channel?

The key here is linearity; a (real-)linear map $\Phi$ defined on all self-adjoint bounded operators on $H$ has a unique extension to all of $B(H)$. This is due to the fact that every $B\in B(H)$ can ...
Frederik vom Ende'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

Finding a succinct representation of a CPTP map

I'm interpreting this as a notation question for $\mathcal{N}$ restricted to act on $\{I, Z\}$. One notation I like is to write $\mathbf{Z}^{\mathbf{b}}:= \bigotimes_{i=1}^k Z_i^{b_i}$, where $\mathbf{...
forky40's user avatar
  • 7,183
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

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