I am trying to understand how decoherence explains the observation of only one possible outcome rather than observing the superposition. When the electron goes through the two slits it interferes with itself in a wave, which makes sense. But when it interacts with the detector, how do we know that the states described by detection at each point on the screen are orthogonal? It feels like there should also be outcomes where the electron is also observed at some combination of these locations.
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There are a few things about what you've written that I don't understand. In particular "one possible outcome rather than observing the superposition" and " outcomes where the electron is also observed at some combination of these locations". Can you clarify your setup and what you mean? – garyp Mar 01 '21 at 18:19
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When the electron hits the detector, we only see it hit one location. We don't observe the electron hitting the detector in a wave (which it was before measurement). I am confused about why the wave necessarily chooses one of these locations rather than collapsing into some collection of points/other pattern. – Jeff Bass Mar 01 '21 at 18:52
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Maybe this sounds too obvious, but the detector is used to measure the location of the electron. It's forced to provide one definite answer, hence it responds with one single location. – A. P. Mar 01 '21 at 20:50
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@A.P. Yeah I'm just operating with assumptions that not everyone agrees with, so it's tough to get an answer. Specifically, I'm trying to work this out within interpretations like many-worlds where the wavefunction is real and all that exists. I'm just trying to work out how the Schrodinger equation naturally evolves in such a complicated system (ie when the electron interacts with the detector) – Jeff Bass Mar 01 '21 at 20:53
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So, your question is basically how observation collapses the wavefunction? – A. P. Mar 01 '21 at 21:03
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@A.P. That helped! Ok I think I'm getting it. – Jeff Bass Mar 01 '21 at 21:57
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When the electron goes through the two slits it interferes with itself in a wave, which makes sense.
This is not correct. The wave nature of the electron is in the wavefunction $Ψ$ modeling it, and the wave is a probability wave = $Ψ^*Ψ$. A single electron leaves a footprint of a particle in the double slit experiment. The accumulation of a lot of electrons with the same boundary conditions shows the wave nature of the crossection/probability. Here is the experiment one electron at a time hitting the screen.
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Each electron leaves a point on the screen, seemingly random in the top frames. It is the accumulation that shows the probability distribution that has wave interference patterns.
But when it interacts with the detector, how do we know that the states described by detection at each point on the screen are orthogonal?
The electron interacts with the atoms in the screen, ionizing them as it passes. It is a much more complicated wavefunction then the one when it scatters through the slits , hitting the screen means new boundary conditions.
It feels like there should also be outcomes where the electron is also observed at some combination of these locations.
The electron is a particle, its wave nature is in the probabiity distributions of the quantum mechanical solution of the scattering problem with its boundary conditions "electron hitting double slits, given distance apart given width "
anna v
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That all makes sense. I guess the confusion is about the boundary conditions when hitting the screen. I don't really buy the idea that the particle is always a particle, so this is just going to be a point of contention. However, my understanding is that if we don't postulate collapse, then the wavefunction would separate into orthogonal states for each of the possible particle locations. I'm just trying to make sense of why these states would be orthogonal and there wouldn't be any other possibilities. – Jeff Bass Mar 01 '21 at 19:32
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