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Bear with me if I present a lack of knowledge - QM is not my field.

There's a common notion in QM that until a particle is observed (measured), its properties are not definite, but rather are spread in a distribution of possible values. it is only upon the act of observing (measuring) that the properties gain a definite value.

My questions are:

  1. If the act of measurement is just the particle interacting with the particles of the detector, then doesn't that imply that particles are almost always observed by all the other particles they interact with? Why is a photon that passes through a double-slit not observed by the slits it's passing through, but by a detector that determines its position after the double- or whichever slit through which it went through in a which-way experiment? Are there interactions that are not considered as observations?

  2. Can a particle be 'unobserved' back into a wave function? Do particles remain 'definite' after observation?

I realize that's more than one question, but an explanation that would help me understand the general concept and the answers by myself is also very much welcome.

Also, please spare me the mathematics, or don't make it the main part of the answer - I'm interested in understanding on a conceptual level.

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    $\begingroup$ Can a particle be 'unobserved' back into a wave function? Do particles remain 'definite' after observation? After the measurement the time evolution of the state of the particle is again governed by the Time-Dependent Schrodinger Equation. $\endgroup$ – Andrei Geanta Jun 13 '18 at 7:14
  • $\begingroup$ Pretty interesting and very related video: youtu.be/8ORLN_KwAgs $\endgroup$ – Gabe12 Jun 13 '18 at 7:20
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    $\begingroup$ physics.stackexchange.com/questions/152906/… $\endgroup$ – Andrei Geanta Jun 13 '18 at 7:45
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    $\begingroup$ "Are there interactions that are not considered as observations?" - Yes, the distributed interactions that support the wave behavior. For example, a reflection from a mirror does not collapse the photon's wave function amazingly despite the fact that the reflected photon is not even the original photon that was absorbed by the mirror, but a new photon re-emitted by the mirror. Refraction (e.g. lenses) also does not collapse the wave function. (2) After the interaction a particle (unless absorbed) continues again as a wave, but a new wave, not the same wave as before the interaction. $\endgroup$ – safesphere Jun 13 '18 at 9:04
  • $\begingroup$ In the case of the mirror, it is elastic scattering, and not absorption-reemission. $\endgroup$ – Árpád Szendrei Jun 13 '18 at 17:57

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