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This course was \$49 per course as verified certified learning outcome - a series of 3 courses + 1 more extra course on edX. Now the course has been taken down; at least no more new enrollments, and MITx Pro is offering it for \$2250 + \$2250 = \$4500. This is 20 times higher for the same course. I see there would be number of other MOOC coming soon. One ...


2

I have some ideas 1) Since $[\mathcal{O}^\dagger_A, \mathcal{O}] = - [\mathcal{O}_A, \mathcal{O}^\dagger]^\dagger$ and $\mathcal{O} = (\mathcal{O}+\mathcal{O}^\dagger)/2 + (\mathcal{O}-\mathcal{O}^\dagger)/2$ you can solve it only for self-adjoint operators $\mathcal{O}=\mathcal{O}^\dagger$. And even more, you can consider only real self-adjoint operators $...


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Try to run this code: from qiskit import IBMQ IBMQ.save_account('your_api_token') And for access: IBMQ.load_account()# without the 's' Hope this helps.


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Based on what I've understood from your question, I think the interpretation of the noisy channel as the quantum algorithm may be causing unnecessary confusion. QEC should be present in the fault tolerant computing paradigm as a background operation capable of guaranteeing that qubits retain coherence of their quantum states for practical amounts of time. As ...


1

These are two separate questions, so I will (try to) answer them separately as well. Concerning the (reduction of) privacy in remote or cloud computing. Without any alterations, the instructions for a quantum computation that is to be run on a remote computer can be seen by that remote computer. That is to say, if you want to conceal the computations that ...


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I don't believe that this is true (at least without further qualifiers). If I'm correctly understanding what you're trying to convey, I think we can essentially simplify this statement to asking whether there can exists a state $|\psi\rangle$ and two projectors $P_1$ and $P_2$ such that $$P_1|\psi\rangle\neq P_2|\psi\rangle$$ (ignoring normalisation for ...


1

We can proceed by applying the parity check matrix $H_2$ for $C_2^\perp$ on the ancilla, which would become $\vert H_2(z'+e_2)\rangle =\vert H_2(z'+u + e_2 -u)\rangle =\vert H_2(e_2-u)\rangle$. Measuring the ancilla gives us $H_2(e_2-u)$. As $u$ is known. we then know $H_2 e_2$, and in turn $e_2$. After correcting the now bit flip $e_2$, we have: $$\dfrac{1}{...


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