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Mar 11, 2022 at 10:29 comment added Andrew Steane suggest you try my answer to this question: physics.stackexchange.com/questions/457908/…; this should clear it up
Mar 11, 2022 at 6:18 answer added user34722 timeline score: 1
Mar 11, 2022 at 5:42 comment added march "the action of an observable operator on a quantum state is a projection into one of the operator's eigenstate". I mean, no, that's not what the action of the operator on the quantum state does. The action of the operator corresponding to the observable has basically nothing to do with the measurement of that observable. You have to define the projection operators onto the eigen-subspaces of the observable separately, and it is the action of those projection operators that makes up the measurement process.
Mar 11, 2022 at 5:30 history edited hodop smith CC BY-SA 4.0
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Mar 11, 2022 at 4:29 history edited hodop smith CC BY-SA 4.0
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Mar 11, 2022 at 4:24 history asked hodop smith CC BY-SA 4.0