Timeline for If the wavefunction of a quantum system is not an eigenfunction of some operator, how do we measure that property?
Current License: CC BY-SA 4.0
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| Oct 10, 2022 at 21:25 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Oct 10, 2022 at 21:22 | comment | added | BioPhysicist | @Akash For Hermitian operators, yes! | |
| Oct 10, 2022 at 16:36 | comment | added | Akash | When you write |c_n|^2=|⟨a_n|ψ⟩|^2, is that only for an orthonormal eigenbasis? Do we know that there is always an orthonormal eigenbasis for an operator? | |
| Nov 4, 2019 at 22:03 | vote | accept | AHMED KRS | ||
| Nov 4, 2019 at 22:02 | comment | added | BioPhysicist | @AHMEDKRS Yes. $|\psi\rangle$ tells us the probability of measuring an eigenvalue of the operator associated with the observable in question. | |
| Nov 4, 2019 at 22:01 | vote | accept | AHMED KRS | ||
| Nov 4, 2019 at 22:03 | |||||
| Nov 4, 2019 at 22:01 | vote | accept | AHMED KRS | ||
| Nov 4, 2019 at 22:01 | |||||
| Nov 4, 2019 at 22:00 | comment | added | AHMED KRS | Yeah, I mean the probability of the measurement. | |
| Nov 4, 2019 at 21:58 | comment | added | BioPhysicist | @AHMEDKRS Kind of. We cannot "get a measurement" in QM. All we can get from $|\psi\rangle$ is the probabilities associated with certain measurements. There is no way to say what the result of a measurement will actually be though. | |
| Nov 4, 2019 at 21:57 | comment | added | AHMED KRS | Yeah, so you are saying that $\psi$ is generally is not an eigenvector of the operator and that when we want to get a measurement of some observable we can expand the wavefunction as a sum of the eigenbasis of that observable? we then use that to get the measurements. | |
| Nov 4, 2019 at 21:07 | history | answered | BioPhysicist | CC BY-SA 4.0 |