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Apr 13, 2017 at 12:39 history edited CommunityBot
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Sep 2, 2015 at 2:26 history tweeted twitter.com/#!/StackPhysics/status/638900738559746048
Sep 1, 2015 at 3:02 comment added Timaeus And this is because you need to get the dynamics of the measurements correct.
Sep 1, 2015 at 3:01 comment added Timaeus In general it's a bad idea to say the square is the probability density. You want to get the correct dynamics which will include eventually getting the frequency of results from repeated interactions with an ensemble of subsystems. But calling it a probability right away can lead to problems if you start to do it too much. For instance if you do it with position then you have to stop you can't do it with position and spin. To get the correlations right then spin had to sometimes be determined by position and by context if you assign position by regular probabilities.
Sep 1, 2015 at 2:28 comment added Gold I imagined that, although the book I'm studying doesn't make it explicit. In that case, one should consider the configuration space $Q$ as we do in classical mechanics and then $\psi : Q\to \mathbb{C}$ will be such that $|\psi|^2$ is the probability density that the system is in the neighborhood of a certain configuration?
Sep 1, 2015 at 2:26 comment added Timaeus Keep on mind that a wavefunction can and often is defined on configuration space which is six dimensional, nine dimensional, twelve dimensional, or more. $\mathbb R^{3n}$ for $n$ particles.
Aug 31, 2015 at 23:20 history edited Qmechanic
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Aug 31, 2015 at 22:53 answer added Javier timeline score: 4
Aug 31, 2015 at 22:44 answer added ACuriousMind timeline score: 6
Aug 31, 2015 at 22:32 history edited Gold CC BY-SA 3.0
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Aug 31, 2015 at 22:27 history asked Gold CC BY-SA 3.0