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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
Nov 4, 2019 at 22:03
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 22:01 vote accept AHMED KRS
Nov 4, 2019 at 22:01
Nov 4, 2019 at 21:57 history edited BioPhysicist CC BY-SA 4.0
added 9 characters in body
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