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Comment by DaveInNYC2 on Rationality Quotes 13 · 2008-09-23T21:18:49.000Z · LW · GW

I always found it strange that the most vocal supporters of flag-burning amendments are military types. Why bother risking your life "for freedom" if you are just going to come home and try to eliminate that freedom?

Comment by DaveInNYC2 on On Being Decoherent · 2008-04-27T18:29:11.000Z · LW · GW

BTW, are you going to cover why the probability of a configuration is the square of its amplitude? Or if that was somehow already answered by the Ebborians, could I get a translation? :)

Comment by DaveInNYC2 on On Being Decoherent · 2008-04-27T18:17:27.000Z · LW · GW

I think this has been the best post so far. I'd like to answer one of my previous questions to make sure I am grokking this; please weigh in if I am off. Here was my question:

Q: Am I correct in assuming that [the amplitude distribution] is independent of (observations, "wave function collapses", or whatever it is when we say that we find a particle at a certain point)?

A: As I suspected, a bit of a Wrong Question. But yes, there is only one amplitude distribution that progresses over time

Q: For example, let's I have a particle that is "probably" going to go in a straight to from x to y, i.e. at each point in time there is a huge bulge in the amplitude distribution at the appropriate point on the line from x to y. If I observe the particle on the opposite side of the moon at some point (i.e. where the amplitude is non-zero, but still tiny), does the particle still have the same probability as before of "jumping" back onto the line from x to y?

A: As was mentioned to me when I first asked this question, the probability of me observing the particle "jump back" is near-zero. When (I think) I realize now is that the reason this is true is that

[Brain being in a state where it remembered particle on the moon 1 second ago] * [Brain being in state where it sees particle "back" on the original line]

is close to zero. (This is of course ignoring the fact that there is no such thing as "this particle" vs "that particle". I'm pretending this is a new class of particle of which there is only one in the universe. Or something).

Comment by DaveInNYC2 on The So-Called Heisenberg Uncertainty Principle · 2008-04-23T18:27:23.000Z · LW · GW

I'm re-going through posts, and a question after reading The Quantum Arena. There you state that if you know the entire amplitude distribution, you can predict what subsequent distributions will be. Am I correct in assuming that this is independent of (observations, "wave function collapses", or whatever it is when we say that we find a particle at a certain point)?

For example, let's I have a particle that is "probably" going to go in a straight to from x to y, i.e. at each point in time there is a huge bulge in the amplitude distribution at the appropriate point on the line from x to y. If I observe the particle on the opposite side of the moon at some point (i.e. where the amplitude is non-zero, but still tiny), does the particle still have the same probability as before of "jumping" back onto the line from x to y?

As I write this, I am starting to suspect that I am asking a Wrong Question. Crap.