Timeline for Why is Gleason's Theorem not enough to obtain Born Rule in Many Worlds Interpretation?
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| May 7, 2017 at 18:38 | comment | added | akhmeteli | @doublefelix: I don't see a significant difference between the two different formulations we use, but "that this type of expression will always work" does not mean it is indeed a probability measure, say, for many-worlds. | |
| May 7, 2017 at 13:18 | comment | added | don't train ai on me | Hmm, I checked the wording of the result (on the last page of Gleason's paper). I'd paraphrase it as "for any probability measure under some assumptions, there is always a density matrix $\rho$ so that the probability is equal to $tr(\rho P)$". This doesn't exclude other, very different expressions which also equal the probability. It does guarantee that this type of expression will always work, though. | |
| May 7, 2017 at 12:34 | comment | added | akhmeteli | @doublefelix: I used the formulation in the Wikipedia that was current at the time I wrote my answer. Do you think that formulation was incorrect? I found it at books.google.com/… as well. | |
| May 7, 2017 at 11:19 | comment | added | don't train ai on me | Hmm, I think the wikipedia for Gleason's Thm, and John Preskill's definition of the theorem in his Quantum Information notes are in disagreement with your comment. Gleason's Theorem is a representation theorem. It says that $tr(\rho P)$ can always represent the probability map from projectors to reals (given a few assumptions). This does not mean it is the only possible representation of such a map. Correct me if I'm wrong. | |
| Feb 12, 2012 at 23:34 | history | answered | akhmeteli | CC BY-SA 3.0 |