Question about the paper A fault-tolerant neutral-atom architecture for universal quantum computation

Quantum Computing is part of Stack Overflow’s open communities: specialist spaces where curiosity is welcome, knowledge is shared freely, and the best answers rise to the top. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more Bring the best of human thought and AI automation together at your work. I've been trying to go through the paper A fault-tolerant neutral-atom architecture for universal quantum computation published in Nature, and I've been bouncing against something, and I was hoping someone could explain what's going on. In the paper, they have (and they repeat this several times throughout the paper) A key experimental upgrade involves non-destructive, spin-resolved qubit readout [emphesis added] My understanding that any measurement of a qubit caused a Quantum State collapse (which limited the Quantum States to be compatible with the measurement), and that could not be called non-destructive; that is, it irretrievably discards part of the Quantum State. They mention that they somehow convert "the atom spin state into position", but as long as position is part of the Quantum State, measuring it still runs into the same issue (and if it is not, that conversion would count as the measurement). What am I missing here? Do they have a different definition of "readout" than what I was assuming (which is moving some information of the Quantum State into the non-Quantum parts of the device)? They do mention that the "readout" was done by a camera, so it would appear that they agreed with my assumption. If I understood correctly, they mean non-destructive in the sense that the atom is still presence at the end of the measurement, which is not true for previous realizations. In the previous realizations, having an atom at the end of the readout would mean you are in the |0⟩|0\rangle state, while not having an atom would put you either in the |1⟩|1\rangle state or you just lost the atom. In the new work, you can differentiate between loss, |0⟩|0\rangle and |1⟩|1\rangle since the measurement does not lead to loss (but rather to movement of the atom). Thanks for contributing an answer to Quantum Computing Stack Exchange! Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. This comment attacks a person or group. 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