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
22 events
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Nov 11, 2019 at 18:00 | history | tweeted | twitter.com/StackPhysics/status/1193951665352847360 | ||
Nov 5, 2019 at 3:21 | history | became hot network question | |||
Nov 4, 2019 at 22:03 | vote | accept | AHMED KRS | ||
Nov 4, 2019 at 22:03 | vote | accept | AHMED KRS | ||
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Nov 4, 2019 at 22:01 | vote | accept | AHMED KRS | ||
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Nov 4, 2019 at 22:01 | vote | accept | AHMED KRS | ||
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Nov 4, 2019 at 22:01 | vote | accept | AHMED KRS | ||
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Nov 4, 2019 at 21:57 | history | edited | BioPhysicist | CC BY-SA 4.0 |
added 9 characters in body
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Nov 4, 2019 at 21:56 | answer | added | ZeroTheHero | timeline score: 3 | |
Nov 4, 2019 at 21:41 | comment | added | BioPhysicist | @AHMEDKRS Please see my answer regarding that point. You don't do $A|\psi\rangle$ to do a measurement. | |
Nov 4, 2019 at 21:36 | comment | added | AHMED KRS | @AaronStevens Yes | |
Nov 4, 2019 at 21:35 | comment | added | MaximusIdeal | @AHMEDKRS This is sort-of the paradox with qm. If the wavefunction is not an eigenstate of an operator, then the corresponding property is not well-defined at that time. However, you can always write the wavefunction as a linear combination of eigenstates (because observable operators are hermitian), so it will instead be a superposition of those eigenstates (until a measurement changes it to a well-defined eigenstate). | |
Nov 4, 2019 at 21:07 | answer | added | BioPhysicist | timeline score: 6 | |
Nov 4, 2019 at 20:57 | comment | added | BioPhysicist | Ah I see. You think that if you want to measure observable $A$ then you calculuate $A|\psi\rangle$? | |
Nov 4, 2019 at 20:55 | comment | added | AHMED KRS | Well, I do not know another method to measure the observable? | |
Nov 4, 2019 at 20:52 | comment | added | BioPhysicist | No, it is not. Which is why I asked my original question. Why do you think the state needs to be an eigenstate of the operator of the observable in order to measure that observable? | |
Nov 4, 2019 at 20:51 | comment | added | AHMED KRS | we've covered this, but is the superposition necessarily an eigenfunction of the operator? | |
Nov 4, 2019 at 20:47 | comment | added | BioPhysicist | The superposition determines the probability of measuring a certain value of the observable. Your QM text/class should cover this. It's a pretty important part of QM. | |
Nov 4, 2019 at 20:42 | comment | added | AHMED KRS | So if it is in a superposition how that is going to change the situation? | |
Nov 4, 2019 at 20:28 | comment | added | BioPhysicist | Why do you think that your system needs to be in a definite state of some observable in order to measure that observable? | |
Nov 4, 2019 at 20:21 | answer | added | Nugi | timeline score: 4 | |
Nov 4, 2019 at 19:16 | history | asked | AHMED KRS | CC BY-SA 4.0 |