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[QUOTE="gentzen, post: 6803414, member: 685984"] Sure. Perhaps I should have written "indistinguishable particles" instead of "identical particles". I will elaborate it for Bosons, so that I can ignore the wavefunction for the equivalence relation. Let the trajectories of the ##n## indistinguishable Bosons be ##(x_1(t), \ldots, x_n(t))\in\mathbb R^{3n}##. We want (at least) the equivalence relation that ##(x_{\pi(1)}(t), \ldots, x_{\pi(n)}(t))## is equivalent to ##(x_1(t), \ldots, x_n(t))## for each permutation ##\pi\in S_n##. This "just works," if the wavefunction is invariant under those permutations. A skeptic might object that we can still use the continuous trajectories to identify a specific particle between different points in time ##t_1## and ##t_2##. Good, but we can prevent that too, by using a "bigger" equivalence relation, on a slightly different space. For example, we can interpret the trajectories as a function ##(x_1, \ldots, x_n)(t) : \mathbb R \to \mathbb R^{3n}## and consider "piecewise constant" permutations ##\pi(t):\mathbb R \to S_n## for the equivalence relation. So ##(x_1, \ldots, x_n)(t)## is "declared" equivalent to ##(x_{\pi(t)(1)}, \ldots, x_{\pi(t)(n)})(t)##. No additional restrictions on the wavefunction are needed (besides those already imposed for the simpler equivalence relation), and now even a specific particle can no longer be identified between different points in time. Of course, a skeptic might have further objections, but they can all be addressed in one way or another, basically because the quotient somehow "miraculously just works" in this case. [/QUOTE]
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