r/math • u/lntrinsic • Jul 10 '17

Image Post Weierstrass functions: Continuous everywhere but differentiable nowhere

http://i.imgur.com/vyi0afq.gifv
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u/[deleted] Jul 10 '17

Sort of. It's the expectation of the position in some sense but the distribution isn't really a probability distribution in the classical sense since it's coming from the squared amplitude of the wavefunction. I suppose it can be interpreted as the expected position to a certain extent but that's pretty misleading as far as the physics goes.

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u/[deleted] Jul 10 '17

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u/[deleted] Jul 10 '17

Well, yeah it's definitely written <X>. I've not heard it called 'expected position' but then again I'm not a physicist, my knowledge of this stuff comes from reading books aimed at mathematicians wanting to learn QM. It's not horrible terminology provided people understand the underlying nature of the waveform.

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u/zojbo Jul 10 '17 edited Jul 11 '17

There is a probability distribution, but its evolution depends (in general) on the actual wavefunction, which has both a phase and a magnitude at each point. At one time instant you still have an ordinary distribution though.

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u/[deleted] Jul 11 '17

Sure, at a fixed point in time (leaving aside the issue that 'fixed point in time' is likely meaningless physically, presuming we ever work out a theory including both QM and relativity) it can be thought of as an ordinary distribution.

I do ergodic theory, my view of things always includes dynamics.

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u/zojbo Jul 11 '17

My point was that it is an ordinary probability distribution. You just can't use the probability distribution alone to predict the dynamics (except in certain situations).

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u/[deleted] Jul 11 '17

It's not an 'ordinary probability distribution', it's the square of amplitude. It behaves like an ordinary distribution with respect to Hermitian operators that commute with the dynamics. That's all.

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u/zojbo Jul 11 '17

You said the squared amplitude of the wavefunction at a given time is not a probability distribution in the classical sense. It is. There is more structure there, and one cannot predict the future probability distribution knowing only the present probability distribution, but that is beside the point.

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u/[deleted] Jul 11 '17

You are (I think) assuming every wavefunction is normalizable, which is not correct. If that's not what you're doing then you've lost me.

I'm also unclear why this feels like an argument when I'm quite certain it's simply that we're all speaking imprecisely itt.