I hear that observation is what causes the wave function collapse, but that doesn't make sense considering that an eye or camera is just a physical system with no particularly special properties. In explanations it say that light causes the collapse if that's the case why do you need an observer? If you shined a light while not in the room of the double silt experiment would it still act as a particle?
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possible duplicate of What counts as "observation" in Schrödinger's Cat, and why are superpositions possible? – John Rennie Feb 05 '14 at 15:33
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I've flagged what I think is a close duplicate of your question. See also How isolated must a system be for it's wave function to be considered not collapsed? If these don't address your question maybe you could edit your question to indicate what specific points you're unsure about. – John Rennie Feb 05 '14 at 15:35
1 Answers
Physically having a human in the room is not necessary to spoil the interference. In the double slit experiment you get interference between particles passing through each hole. However this interference is spoiled if you put any sort of "indicator" on each hole so you can check which hole the particles are going through (for concreteness lets consider electrons).
Microscopically, this "indicator" can be two streams of very low energy photons that scatter off the electrons, one for each hole. Then to know which hole the electrons went through you could in principle check where a photon was scattered. It does not matter that there is an actual observer in the room. However, it does matter that in principle you could use the photons to measure where the electron is. Furthermore, to fully spoil the interference pattern, the photon must be produced often enough that each electron will interact with a photon.
That being said, if the light was on in a room you could still see interference since the probability of an electron scattering off of the photons in the room is low. This will occur occasionally and smear your interference pattern a bit, however you will still see interference.

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