- #1

andresB

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Any thoughts?

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- Thread starter andresB
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- #1

andresB

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Any thoughts?

- #2

atyy

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The quick answer is that one doesn't need collapse unless one is doing another measurement after the first. So we can just declare the measurement done, and the particle destroyed after that.

However, the collapse rule is actually more general, and doesn't have to be a projection onto an eigenvector of the observable. The heuristic way to see this is that what you consider a measurement is subjective, so you can collapse it onto the eigenvector and rotate it, and call that process the measurement.

OK, but that still doesn't help with continuous variables, since the eigenvector itself is not a legitimate wave function. For continuous variables the collapse is defined by Eq 3 of http://arxiv.org/abs/0706.3526. Remarkably, Ozawa discovered that there is a measurement model that will give a sharp position measurement, which is summarized in section 2.3.2 of that paper.

However, the collapse rule is actually more general, and doesn't have to be a projection onto an eigenvector of the observable. The heuristic way to see this is that what you consider a measurement is subjective, so you can collapse it onto the eigenvector and rotate it, and call that process the measurement.

OK, but that still doesn't help with continuous variables, since the eigenvector itself is not a legitimate wave function. For continuous variables the collapse is defined by Eq 3 of http://arxiv.org/abs/0706.3526. Remarkably, Ozawa discovered that there is a measurement model that will give a sharp position measurement, which is summarized in section 2.3.2 of that paper.

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- #3

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That means that you cannot measure position with infinite precision, but you can measure position with an arbitrarily good finite precision. For all practical purposes, this is the same.

Any thoughts?

- #4

bhobba

Mentor

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That means that you cannot measure position with infinite precision, but you can measure position with an arbitrarily good finite precision. For all practical purposes, this is the same.

This is seen by the fact a state with perfect position is a Dirac Delta Function - which isn't really a function, and certainly doesn't belong to a Hilbert space.

Thanks

Bill

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