From: Stuart Armstrong (email@example.com)
Date: Tue Feb 10 2009 - 03:47:47 MST
Nice derivation - but I've seen it before. Seems like a general
procedure that people keep on rediscovering.
And the probability is NOT the square of the (complex valued) wave
function, but the modulus squared (or F(x) times the conjugate of
F(x)). Then you need to add a remark to get rid of the non-real phase
changes (such as multiplication by i) - simply note that if you do the
interchange twice, then you return to the original system, so the
phase change has to square to one.
2009/2/9 John K Clark <firstname.lastname@example.org>:
> From my post to the Extropian list:
> experimentally we can't measure the quantum wave function F(x) of a
> particle, we can only measure the intensity (square) of the wave
> [F(x)]^2 because that's a probability and probability we can measure.
> Let's consider a very simple system with lots of space but only 2
> in it. P(x) is the probability of finding two particles x distance
> and we know that probability is the square of the wave function,
> so P(x) =[F(x)]^2. Now let's exchange the position of the particles in
> the system, the distance between them was x1 - x2 = x but is
> now x2 - x1 = -x.
> The Identity Of Indiscernibles tells us that because the two particles
> the same, no measurable change has been made, no change in probability,
> P(x) = P(-x). Probability is just the square of the wave function so
> [ F(x) ]^2 = [F(-x)]^2 . From this we can tell that the Quantum
> wave function can be either an even function, F(x) = +F(-x), or an odd
> function, F(x) = -F(-x). Either type of function would work in our
> probability equation because the square of minus 1 is equal to the
> of plus 1. It turns out both solutions have physical significance,
> with integer spin, bosons, have even wave functions, particles with half
> integer spin, fermions, have odd wave functions.
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