= 27d567bc2d48d9f40e9ccc20ea370b15)
nshepperd comments on The Quantum Arena - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (71)
This series is great. But, I'm having a little trouble understanding the fourth diagram, the one with the folded configuration space.
I sort of get it: the original configuration space distinguished between two particles, which is wrong, so in reality only half of the configuration space's area matters when it comes to information. But I don't get how that means you delete the probability from half of the space. Why is it wrong to make the space symmetrical across the diagonal line? It seems a little arbitrary to me; is there a physical reason, or is this a standard thing to do?
Also, I don't get why "this identity cuts down the size of a 2-particle configuration space by 1/2, cuts down the size of a 3-particle configuration space by 1/6, and so on." What's the relationship from 1/2 to 1/6? What comes after that? Why isn't it 1/2 to 1/4 to 1/8?
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)?