nshepperd comments on The Quantum Arena - Less Wrong
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Comments (71)
For a three particle configuration space, you could imagine tagging each of the three particles as "particle A", "particle B" and "particle C". So for a classical configuration with a particle at each of the positions X, Y and Z, you would have six ways of assigning the particles to the positions -- if we represent them as (particle at X, particle at Y, particle at Z), we've got (A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A). In general, for n particles, the number of ways is just the factorial n! (the number of permutations of n elements), which you may have guessed by now.
But of course in quantum configuration space there's only one way of having "particles at X, Y and Z", so we cut down the space by 1/(n!).
Thanks, that explains the 1/2, 1/6, etc. thing. So 1/24 is indeed next.
I still don't get the folding thing (vs. making the picture diagonally symmetrical) very much, but I kind of get it, so I'll leave it be.
Maybe you don't just make the picture diagonally symmetrical because then if you would integrate over all the probabilities you would get 200% (because you represent every situation twice)?