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Nov 10, 2017 at 19:54 comment added Sunyam @JakobLindemann Please see the addendum to my answer for the details you are looking for in BySymmetry's answer.
Nov 10, 2017 at 16:40 comment added Jakob Lindemann Thanks for your answer. This makes sense in and of itself, and I wish I had found a textbook that states this as clearly as you have here. That said, I'm still not clear on how this relates to Sunyam's answer regarding the more "obvious" explanation in terms of integration against the normalised probability distribution. That is, is there a straightforward way to show that this functional integral is exactly that set forth by Sunyam for the canonical ensemble?
Nov 10, 2017 at 15:04 history answered By Symmetry CC BY-SA 3.0