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This is wholly analogous to the evanescent optical field that arises in the classically (i.e. computed by raytracing) forbidden region beyond a totally internally reflecting interface between two optical mediums. I analyse this situation in my answer here and there is also a great plot of the situation in Ruslan's answer here. Let's think of a 1D barrier ...

3

Two cents from an experimentalist. It is always good to keep in mind that a wavefunction for a real particle in the lab is a solution of Schrodinger's equation with specific boundary conditions given by the experimental setup that makes the measurement. Every measurement changes the boundary conditions for the solution that describes the particle's ...

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The electron doesn't get destroyed when you measure it (though photons usually do), but its wavefunction doesn't go back to how it was before. Instead it gets a new wavefunction, different from the old one. If you measured the position of the electron, this new wavefunction will be a delta function (a single infinitely sharp spike) centred at the position ...

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Assuming wave-function collapse is correct (which can be a relatively hefty philosophical claim in some circles), then think of measurement this way: When you measure an observable on a system, you collapse the wave-function of the system into a Dirac delta function in the eigenbasis for that observable. If you measure position, you get a delta function in ...

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In practice, the apparatus measuring the spin should be localized somewhere in space (it cannot fill the whole universe!) and this fact implies that you always make a measurement of position (actually very rough in general), even if you are measuring the spin. Suppose that $\Omega \subset R^3$ is the bounded region in $R^3$ where the apparatus is localized. ...

0

Let's start off by removing the restriction of computational resources such that we're not limited by computing power and by finite precisions. Let's also use the word exact to mean absolute certainty (ie. probability is precisely 1) about a quantity. Take a real group of particles at an initial state. We may or may not be able to derive a set of governing ...

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As Emilio pointed out, the uncertainty principle is not a limiting factor. However, as for simulating or calculating future states, this is not really generally possible for classical systems, because of chaotic behaviour.

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The Uncertainty Principle will never, as far as we know, prevent you from simulating any physical system. The reason for this is that quantum mechanics is - except for that little problem with measurements - completely deterministic. To be more precise, say you want to simulate a given system within quantum mechanics. You begin by describing your ...

2

My reading of the electron photon experiment is that intrinsic uncertainty enters the problem by limiting the resolving power of the photon. In other words, the electron is along for the ride, and perhaps historically it was chosen because it is such a simple system. But the recoil of the electron seems to confuse the issue. Instead of a free electron, we ...

2

Yes, the experiment is oversimplified, because the uncertainty principle is not about "disturbance through measurement". Although that's what Heisenberg said (one of the things he said), it turned out you can't interpret it that way in a very rigorous sense. Whether there is something like "disturbance through measurement" that gives rise to an uncertainty ...

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Very intuitive. No maths. There is an excited state with a symmetrical probability distribution and no e/m dipole moment. There is a ground state (or less excited state) also with no dipole moment. There is a tiny probability that the excited state electron will be in the ground state that allows both states to be present at the same time producing a finite ...

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