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Is it possible to ever actually measure something so precisely that it actually collapses to a pure state, or do we really just get arbitrarily close? If a wavefunction never actually collapses to a pure state, why do we learn about the wavefunction collapsing to an eigenstate?

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    $\begingroup$ If a mass point does not exist, why do we learn about mass points? $\endgroup$ – Sebastian Riese Aug 20 at 19:16
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    $\begingroup$ 1. "Pure state" has a technical meaning as opposed to a "mixed state", which are represented by density matrices (as opposed to single state vectors for pure states). You probably do not mean that here. 2. Stern-Gerlach experiments where you detect clear spin up/spin down states seem like very straightforward examples of results of measurements that are pure eigenstates of the corresponding measured spin operator. Is that somehow insufficient? $\endgroup$ – ACuriousMind Aug 20 at 19:18
  • $\begingroup$ @SebastianRiese But it just seems so strange that I've read all these books that say "the wavefunction collapses to an eigenstate" without any mention of the fact that this isn't physical. When we talk about mass points there's usually a disclaimer. Well if your answer is that it's just a convenient abstraction then it answers my main question. $\endgroup$ – Jeff Bass Aug 20 at 19:18
  • $\begingroup$ @ACuriousMind No, that one makes sense since there are only two possibilities, but I've definitely been getting hung up on position measurements due to the continuous spectrum. I think Valter hit on what I was really asking by saying that the instrument's accuracy must be within the difference of two discrete states in order to obtain a pure state. $\endgroup$ – Jeff Bass Aug 20 at 21:29
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    $\begingroup$ @JeffBass In practice, measurements don't yield one position eigenstate, but a narrow superposition. $\endgroup$ – alexchandel Oct 31 at 18:40
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If you measure an observable with discrete spectrum (or a finite set of observables whose set of discrete eigenvalues determins a single vector state) and the difference between two consecutive eigenvalues is larger than the instruments accuracy, then you always obtain a pure state as an outcome of the measurement. I stress that this does not require an infinite precision of the instruments.

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  • $\begingroup$ And if the accuracy is not good enough the wavefunction will be a mixed state of all the eigenstates that it could still be?...also is a mixed state the same as a superposition? $\endgroup$ – Jeff Bass Aug 20 at 21:30
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    $\begingroup$ In principle, the state should be pure also in that case: a coherent superposition of various eigenststes according to the projection postulate. A mixed state is obtained when you measure as said a number of identical systems equally preparated, obtaining different outcomes according to QM; and then you collect all together the measured systems. Every element of this ensemble stays in the same mixed state... $\endgroup$ – Valter Moretti Aug 21 at 16:11

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