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2 votes

What are the possible channels preserving purity of all input states?

There are no other examples. Consider the Kraus representation $\Phi(X)=\sum_a K_a X K_a^\dagger$. If $\Phi_f(\mathbb{P}_\psi)=\mathbb{P}_{f(\psi)}$ is pure then all $K_a \mathbb{P}_\psi K_a^\dagger = ...
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2 votes

What are the possible channels preserving purity of all input states?

A (partial) answer can be found in 'Transformations on tensor product spaces' of M Marcus, BN Moyls (1959), where they show that any map of matrices to itselves $f: M_n\to M_n$ that leaves the rank-1 ...
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3 votes

What is the adjoint of the complementary channel?

Let us first suppose more generally that we have a map defined as $$ \Psi(X) = \sum_k \langle B_k, X \rangle A_k $$ for all $X$. The adjoint mapping $\Psi^{\ast}$ must satisfy \begin{multline} \langle ...
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4 votes
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Why is $\Phi\otimes \operatorname{Id}_n$ being positive on maximally entangled states sufficient to know that $\Phi$ is CP?

One alternative argument would be as follows: If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\vert\chi\rangle$ such that $\Phi\...
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1 vote

Why is $\Phi\otimes \operatorname{Id}_n$ being positive on maximally entangled states sufficient to know that $\Phi$ is CP?

You can skip point 3 since $\Phi(X) = AXA^\dagger$ is a CP map and a sum of CP maps is again CP. The map $\Phi$ is CP because $(\Phi \otimes {\rm Id})(Z)= (A\otimes I)Z(A\otimes I)^\dagger$ – clearly ...
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2 votes

Why does code switching not allow for universal fault-tolerant quantum computation?

One reference is here. The authors switch between the 7-qubit Steane code and 15-qubit Reed-Muller code. Clifford group operators can be performed in the Steane code, and the T-gate in the 15-qubit ...
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0 votes

Qiskit library.gaussian() does not accept parametric expression

Just change library.gaussian to become library.Gaussian with capital G. Qiskit Pulse contains both ...
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1 vote
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Filtering operation is trace decreasing?

Yes. Generally you should think of filtering as being a measurement. You're only describing the effect of one measurement outcome. There's generally a second one such that the net effect is trace ...
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1 vote

How to comparing a quantum channel with a unitary?

I found the following two works solving this problem. They selected several reference states and took the HS product between the final states evolved by the target unitary and the quantum channel. ...
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3 votes
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Can any rank-$n$ POVM be realized as a rank-one POVM?

Is this an exercise? The answer is yes, you can do it, but I'm not going to tell you what the equivalent rank-1 POVM is. In general, any higher-rank POVM is equivalent to a rank-1 POVM if you relabel ...
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-1 votes

Can any rank-$n$ POVM be realized as a rank-one POVM?

I originally intended this to be a comment, but it got too long for a comment. I think it depends on what you want to do with the POVM, if you are more interested in maximising the amount of ...
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1 vote
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Characterise, via Naimark's theorem, the POVM corresponding to a PVM in a dilated space

You could define $$ V_F^S u \equiv \sum_a |a\rangle\otimes (S^a\sqrt{F^a}\,u), \qquad u \in\mathcal X, $$ where $S^a$ is any unitary. Then $V_F^S$ is also an isometry and $$ F^a=(V_F^S)^\dagger (\...
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