## Posts

## Comments

**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?

**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? :)

**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).

**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.