2

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.

Qmechanic
  • 201,751
Yuval Weissler
  • 1,538
  • 2
    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. – Nemo Jun 13 '18 at 07:14
  • Pretty interesting and very related video: https://youtu.be/8ORLN_KwAgs – Gabe12 Jun 13 '18 at 07:20
  • 3
    https://physics.stackexchange.com/questions/152906/simple-example-showing-why-measurement-interaction-are-different – Nemo Jun 13 '18 at 07:45
  • 1
    "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. – safesphere Jun 13 '18 at 09:04
  • In the case of the mirror, it is elastic scattering, and not absorption-reemission. – Árpád Szendrei Jun 13 '18 at 17:57
  • @ÁrpádSzendrei Please use the @ tag when replying or referring to someone. Otherwise users don't get notified. Photons are massless particles that cannot scatter or change direction in a flat space. For example, in Compton scattering, the photon is actually absorbed and re-emitted: https://upload.wikimedia.org/wikipedia/commons/1/10/Compton_Scattering_Feynman_Diagram.png – safesphere Jun 15 '18 at 04:47
  • @safesphere well I am sorry buy I will have to disagree. In case of a mirror, it is elastic scattering and there should be no absorption-reemission. It is a specular reflection. See Anna V's answer here: https://physics.stackexchange.com/questions/35177/what-happens-when-a-photon-hits-a-mirror – Árpád Szendrei Jun 15 '18 at 05:17
  • @ÁrpádSzendrei I am not sure what you are disagreeing with. Both Anna and John stated there that the photon is absorbed and re-emitted in a mirror. Plus this is a direct quote from your own answer there, "the individual photons are absorbed, and then a 'new' photon is re-emitted, with the same properties, except for the direction which is in a normal mirror opposite". – safesphere Jun 15 '18 at 06:13
  • @safesphere this is not true. Anna V, who is the expert in this field, says in the comment to John's answer: "John, please look at this again, as it is a chosen answer. is not correct for a mirror. Absorption and re-emission would change the phases (the reemitting source would have random direction) and no images would be transmitted to the eye , to call it a mirror. It has to be elastic scattering for a mirror " – Árpád Szendrei Jun 15 '18 at 16:08
  • In her answer she says it is Rayleigh scattering. She says if it is absorbed, and then she says it will be different from a mirror. That is not saying it will be absorbed. It is only saying that if it will be absorbed, the mirror image will not be built. Because if it will be absorbed and re-emitted, the that photon will change lose energy and change frequency, which is not a mirror image. "then a re-emitted photon can change both direction and energy with respect to the originating one, and the originating one loses energy, i.e. changes frequency." – Árpád Szendrei Jun 15 '18 at 16:15
  • " If it is reflected, it of course goes with velocity c (as all photons) whatever its direction (elastic scattering means only change of direction and not energy)." – Árpád Szendrei Jun 15 '18 at 16:16
  • And to my answer there, that is what I said, and then they corrected me later. – Árpád Szendrei Jun 15 '18 at 16:17
  • @ÁrpádSzendrei I believe your understanding is incorrect, but you are entitled to your opinion. A single photon reflection in the mirror is a random re-emission. There is no such a thing as a mirror for a single photon, unless you count many photons one after another. A mirror reflection is a collective process described by the wave behavior. Massless particles don't experience time, so nothing can happen to them in flight. They simply go straight from the emission to the absorption by massive particles. No decay, no scattering, nothing. Everything else is a re-emission of a new particle. – safesphere Jun 15 '18 at 16:27

0 Answers0