Title: quantum mechanics - Observer effect, do this mean literally someone or just any interaction with other matter? - Physics Stack Exchange
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Observer effect, do this mean literally someone or just any interaction with other matter?
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Content: However, any attempt to try to watch it going through both slits at once will require something to interact with that photon, whether that be a human observer or simply the matter that makes up a piece of apparatus. That interaction causes the photon's wavefunction to collapse, and the quantum effect of the two slit defraction to cease. (the uncertainty principle by contrast is the same whether or not the photon is observed)
It is interesting to apply this explanation to Schrodinger's Cat. The radioactive decay that is to trigger the release of cyanide and kill the cat can exist in two states only as long as it interacts with nothing else. There is no need for a human observer, and so no paradox - the wavefunction of the radioactive decay collapses when it interacts with the detector and the cat in the closed box, meaning that the cat is already either alive or dead when the box is opened. Share
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answered Aug 9 '13 at 18:42
Gwilymino
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This "observation" is actually just a more simple way to state the uncertainty principle. For a one dimensional case involving momentum and position it would be as follows. Δ x Δ p ≥ h
This means that if we know the position exactly then we don't know the momentum. For energy and time it would be
Δ E Δ t ≥ ℏ 2
Basically what these mean is that there is a certain amount of maximum precision that we can make. If we knew the energy exactly ( Δ E = 0) then we know that Δ t is AT LEAST ℏ 2.