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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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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
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Action of a CPTP map on Identity

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

Are $n$-qubit Pauli channels $\mathcal E(\rho)=\sum_j p_j P_j \rho P_j$ invertible?

TL;DR: A Pauli channel has a mathematical inverse if and only if it doesn't vanish on any Pauli operator. The inverse is physical if and only if the channel is unitary. The former follows from ...
Adam Zalcman's user avatar
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3 votes
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How to extract probabilities from Kraus representation?

We can indeed rewrite $\mathcal{E}(\rho)=\sum_iK_i\rho K_i^\dagger$ as $\mathcal{E}(\rho)=\sum_ip(i)\rho_i$ by setting $p(i):=\mathrm{tr}(K_i\rho K_i^\dagger)$ and $\rho_i:=\frac{K_i\rho K_i^\dagger}{...
Adam Zalcman's user avatar
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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
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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
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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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
1 vote

Unital channel which is not 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
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1 vote

How does the spectral decomposition of the Choi operator relate to Kraus operators?

This paper is a nice reference on this topic, and presents the results algebraically and graphically
QuantumL's user avatar

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