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5

As far as I understand it, your claim is not per se about measurement but more like "if I prepare a particle in an eigenstate of either position or momentum, doesn't it mean that I have product of uncertainties $0.\infty$?" isn't it? One first problem with this claim as it is, is that the corresponding "wave functions" of these states are not ...


6

When position is measured, the uncertainty of the resulting delta spike's position is 0 This notion is the root of the problem. Quantum states which are actually eigenstates of the position operator are mathematically pathological and also completely unphysical. Some math tools Consider a one dimensional system. Suppose $\{|x\rangle \}$ is an ...


1

You cannot measure "where the particle is not at" in this way. When you send a photon it either interacts with the electron, collapsing its wave function, or it doesn't. If it interacts with the electron, then you can say, with some certainty related to the photon energy, where the electron was. But if not, there's nothing you can say about it. It does ...


1

Furthermore, if there are two particles with similar wave functions (some probability for either to be in the tube), then we will collapse them both. No. Quantum mechanical solutions are exact and deterministic. Their square determines completely the probabilities of observation and each solution has definite boundary conditions that generate it as a ...


0

I am still trying to wrap my head around this experiment. A photon has properties such as frequency and intensity and momentum and direction, even when the observer instrument has "collapsed" the wave function. So a free photon is definitely always a wave (not solid). The detector's measurements tell us that both the "collapsed" photon and the "wave" ...



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