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A completely mixed state is a statistical mixture with no interference terms, and (QMD, McMahon, pg 229):

$$\rho = \dfrac{1}{n}I$$

$$Tr(\rho^2) = \dfrac{1}{n}$$

Are black hole quantum states completely mixed, meaning that they each have the same probability of being observed if we could in fact observe a black hole?

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  • $\begingroup$ Your question isn't specific for black holes and it depends whether one considers a microcanonical or canonical ensemble. In general, a black hole with well-defined enough geometry also has a sharp enough temperature, so one may write $\rho$ as a thermal density matrix but there's still the freedom to consider the different ensembles. Their macroscopic consequences are identical in the large black hole limit - the thermodynamic limit. $\endgroup$ – Luboš Motl May 23 '11 at 9:20
  • $\begingroup$ Apologies, I updated the question in the interim, I am not sure if it is phrased the way I want it yet, I am trying to see if there is a simple statement about probabilities of black hole states. I might edit this later. $\endgroup$ – Unassuminglymeek May 23 '11 at 9:32
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    $\begingroup$ The exact microstate of a black hole is going to reflect the dynamics by which it was formed. Assuming that black hole quantum mechanics behaves like every other kind of quantum mechanics we know about, then if we knew enough about the black hole formation, and we were able to measure the environment completely, then the black hole would be in a pure state. $\endgroup$ – Peter Shor Aug 23 '11 at 15:01
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All microstates with the same energy. Otherwise, we ought to have a Gibbs distribution $e^{-\beta E}$.

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  • $\begingroup$ I am not sure I like the question I posted yet, I will have to try to refine later. $\endgroup$ – Unassuminglymeek May 23 '11 at 9:33

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