# Why Is a star a Pure state?

I am reading some papers about black hole complementarity (Samir D. Mathur. The information paradox: conﬂicts and resolutions. Proceedings for Lepton-Photon 2011 (expanded). arXiv:1201.2079 [hep-th].) and I found the following sentence;

Thus an initial pure state (the gas making up the collapsing star) evolves to a black hole, and after evaporation, into something that can only be described by a mixed state.

How is a star a pure state? Isn't it a thermal state and as a consequence is mixed? Isnt that why the light coming from the sun is unpolarized?

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Do you have a reference for that sentence? – Chris White Mar 1 at 7:23

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– twistor59 Mar 1 at 13:07

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The star is represented by a pure state in the sense that the quantum mechanical state of all its components is completely knowable in principle. The paragraph linked (page 7) argues why this has then been treated as evolving into a mixed state - you throw away some information. – twistor59 Mar 1 at 13:13

## 1 Answer

As twistor59 already correctly wrote, a star and every other physical system may be found in a pure state in principle. When a physical system is in a pure state, it just means that we're assuming the maximum possible knowledge about the system – e.g. we know all the eigenvalues of a complete set of commuting observables.

If a physical system is in a mixed (non-pure) state, it just means that our knowledge about the system is smaller than the maximum one and we're creating a probabilistic mixture of pure states – which are mixed in a way that is fully analogous to probabilistic distributions in classical statistical physics.

In quantum mechanics, predictions are always probabilistic but for any pure state, the role for the probabilities is "minimized". Pure states are the closest objects somewhat analogous to "points in the phase space" in classical physics.

In the black hole information puzzle research and many other considerations, we're considering the initial state of the matter to be in a pure state and we can do so without a loss of generality because the evolution of any mixed state is completely determined by the evolution of pure states that belong to a basis of the Hilbert space (space of possible pure states). The previous sentence, easily proved from the evolution equations for density matrices and pure states, is just the quantum counterpart of the classical statement that it's enough to investigate the behavior of well-defined points in the phase space (pure classical states) and the behavior of all probabilistic mixtures is then determined by purely statistical (Hamiltonian-independent) formulae.

A star of a certain temperature may be described by a mixed state but in principle, we may also assume that the actual state of the star is a pure state that just "resembles" the mixed state – a typical pure state in the ensemble, for example – and we may study the behavior of these pure states separately. Indeed, that's necessary if we want the potential information loss to be seen at all. The information loss was the apparent conclusion by Hawking in the 1970s that a pure state will ultimately evolve into a mixed one. There has never been a doubt that a mixed state evolves into a mixed state but the problem was that even a pure state seemed not to be willing to stay pure. This conclusion was later shown to be an artifact of the perturbative approximations in quantum gravity. The full theory of quantum gravity evolves pure states into pure states of the final Hawking radiation.

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Could you give references for your final comments? – Emilio Pisanty Mar 1 at 14:57
Thank you for the answer. I think it answers my doubt. I would also be interested in the reference for your last sentence. – Super Frog Mar 1 at 16:16
Dear Emilio, it's what the thousands of papers on information loss paradox are all about. See some examples at scholar.google.cz/… - Check that all papers allowing pure-to-mixed evolution are before 1995 or surely before 1997 A focused search on newer papers, AdS/CFT, saying that the evolution is unitary, e.g. scholar.google.cz/… – Luboš Motl Mar 1 at 19:03
– Luboš Motl Mar 1 at 19:05

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Dear @lurscher, a pure (zero von Neumann entropy) state evolves into a pure (zero von Neumann entropy) state. This is the unitarity of the evolution. We ascribe a finite entropy proportional to the area to a black hole but "black hole" either means a mixed state here or the entropy is defined as $\ln N$ where $N$ is the number of macroscopically indistinguishable states from a given microstate, not the von Neumann entropy. A pure black hole (or any other pure) microstate of course carries no von Neumann entropy and no sane physicist has ever claimed otherwise. – Luboš Motl Mar 2 at 7:26
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