From what I saw, it seems that if theoretically you know the current states of a system (which seems impossible), you can predict its future wave function. But since there are wave function collapses, (or the world splitting into infinite many worlds, or hidden variables in pilot wave theory), even if you know the current states of a system, you can't derive the past based on it. Is there such asymmetry? If so, why is future deterministic but not the past?
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$\begingroup$ A deterministic model produces the same output from the same initial state. After the first measurement, the model is no longer deterministic. In general, quantum mechanical models aren't considered to be deterministic. $\endgroup$– Cinaed SimsonCommented Dec 12, 2019 at 21:05
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$\begingroup$ @CinaedSimson the definition of 'measurement' is kind of confusing though, it seems to be defined based on a observer's point of view. But if the observer is included in the system, the wavefunction of the whole system wouldn't be changed? $\endgroup$– zesCommented Dec 13, 2019 at 23:03
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3 Answers
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Neither the past nor the future can be determined by a measurement now. A good example is the famous two-slit interferometer. We can measure the location of a photon detection event in the interference pattern, but that measurement doesn't tell us which slit the photon went through on its way to the detector. If we trace the wavefunction backwards in time, it goes through both slits. So, quantum mechanically, there are multiple possible pasts just as there are multiple possible futures.
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$\begingroup$ Yeah, but it still seems asymmetric though,by 'the current states', I meant all the parameters of the wavefuction at the moment (I don't think it's knowable), but given that, one can calculate how that wavefunction would evolve in the future but would not know if that wavefunction is a result of 'wavefunction collapse' some time in the past. There's probably 'wavefunction collapses' in the future too, I don't know what a collapse is in physics. $\endgroup$– zesCommented Dec 13, 2019 at 22:57
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$\begingroup$ Don't feel alone. "Wavefunction collapse" has been a point of debate, contention, and confusion among physicists for a very long time. $\endgroup$ Commented Dec 13, 2019 at 23:08
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A quantum system evolves deterministically only between measurements*. So if you know the state of a system you can determine its past and future states providing no measurement occurs during the time interval you are concerned with. A measurement is a non-deterministic process where “collapse” occurs in the Copenhagen interpretation.
In short, determinism can only be used to give the state of a system in the time interval between its most recent past measurement and its closest future measurement.
*The definition what “measurement” actually means is an open question
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$\begingroup$ Thank you. I don't know what the difference between a 'measurement' and an 'interaction' is. Every particle is constantly interacting with all other things in the observable universe but this does not seem to automatically constitute a 'measurement'. Maybe it is sometimes the force magnitude or the energy intensity of the interaction which might help measure things more precisely? So say if you use a high energy photon to check if a particle is at a certain region, that would 'collapse' the wave function more than if you use some weak gravitational wave on it? $\endgroup$– zesCommented Dec 12, 2019 at 6:25
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My understanding is that we can determine what happened in the past via use of measurements but for the future we can only use probabilities as a means of trying to determine what is undetermined
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$\begingroup$ I don't think you can measure the wavefunction of a thing? Maybe from repetitive measurements of electrons emitting from the same electron emitter you can approximate the wavefunction of the electrons being emitted? $\endgroup$– zesCommented Dec 13, 2019 at 23:01
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$\begingroup$ By doing those repetitive measurements you will be able to determine probabilities of where you will detect the electrons. You might detect 20% arriving at point A, 37% at point B and so on. Then you can begin to predict with some certainty but never 100% certainty. $\endgroup$– WookieCommented Dec 14, 2019 at 9:21