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A friend was explaining Black Hole Complementarity to me, and at one point he said that to get a (horrendously) mixed quantum state, i.e. a thermal density matrix without a heat bath, one takes a maximally entangled pure state and partial trace. And that's how one would end up getting Hawking radiation. Also, at infinity, there's a thermal density matrix.

I am somewhat confused. I don't understand what is the partial trace for and why is it (necessarily) used in the first place? Is it because we want to restrict our attention to a subspace of the Hilbert space, or in other words when we have to ask about entanglement and subsystems?

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  • $\begingroup$ I've removed a nonconstructive comment discussion. Please keep in mind that comments are meant for suggesting improvements and requesting clarification. $\endgroup$
    – David Z
    Commented Apr 26, 2016 at 8:37
  • $\begingroup$ @DavidZ: I did suggest an improvement: the OP needs to think about the physicality of the posed problem. The remarkably simple answer is that the problem is unphysical to begin with. Math can't give us the right answer when we are asking the wrong question. $\endgroup$
    – CuriousOne
    Commented Apr 26, 2016 at 8:50
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    $\begingroup$ That's not a suggestion for improvement. $\endgroup$
    – David Z
    Commented Apr 26, 2016 at 9:05
  • $\begingroup$ I don't think you need a question about black hole complementarity. You need to ask a question about entanglement and partial trace. $\endgroup$
    – MBN
    Commented Apr 26, 2016 at 12:10
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – MBN
    Commented Apr 26, 2016 at 12:10

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Suppose you have two systems $S_1,S_2$ with Hilbert spaces $H_1,H_2$ with a density matrix $\rho$ on $H_1\otimes H_2$. The partial trace of $\rho$ over the Hilbert space of one of the systems, $H_1$ say gives you a reduced density matrix $\rho_2$. The reduced density matrix predicts the expectation values of all the measurements you can conduct on $S_2$ alone. It does not predict expectation values of measurements on $S_1$, or the expectation values of measurements on the joint system.

In this case, the universe outside the black hole supposedly loses access to information about what falls into the black hole. So the information available to those outside the black hole is described by a reduced density matrix. Whether such loss actually happens is unknown.

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  • $\begingroup$ And I guess the crux of the problem is that after the black hole has completely evaporated, this reduced density matrix is all that you have left. $\endgroup$
    – Rococo
    Commented Apr 26, 2016 at 15:00
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    $\begingroup$ Possibly helpful: backreaction.blogspot.com/2015/04/… $\endgroup$
    – Rococo
    Commented Apr 26, 2016 at 15:06
  • $\begingroup$ The assumption that a black hole doesn't "communicate" with its surrounding, at all, is false, though. LIGO has detected quite an impressive signal from just such a "communication", three solar masses worth of gravitational waves. Anything that falls into a black hole will generate a smaller, but non-zero signal of this kind. $\endgroup$
    – CuriousOne
    Commented Apr 26, 2016 at 18:05
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Since everybody seems to need the kid who cries that the emperor has no clothes, I am more than happy to make the same statement in an answer: the question posed by the black hole complementarity paradox is unphysical.

Information is always lost in any physical systems. Thermodynamics is about nothing else than information loss. Whether it's melting ice cubes losing their shape, a heart on the foam of a cup of latte disappearing by stirring or the initial conditions in a gravitating n-body problem getting tangled to absurdity - we can never get the full information about the past back.

Let me repeat this: that is normal and one can easily deduce it from special relativity: outgoing radiation is leaving a localized system at the speed of light, which, as we hopefully all accept, can not be caught up with. Once thermal radiation is gone, it's gone, and it takes part of the system state with it. Even if we could do the inverse reconstruction of the dynamic (which, as we know from classical mechanics, is not even possible for any but a handful of trivial Hamiltonian systems), we would already be lacking the necessary ingredients for this calculation: the lost heat has destroyed any chance of a full reversal.

Why some theoreticians of otherwise enormous intellect have taken offense with the normal modus operandi of nature is a real mystery, which I admit. The suggestion that black holes are the only information conserving devices in the universe is, on the other hand, very strange and I do not see a single shred of physical motivation for it, let alone any chance of testing this claim. And with that alone the question disappears, by definition, from the pages of science.

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  • $\begingroup$ But that wasn't my question in the first place. I asked why do we need to take the trace in the first place. The question was not even if the information will be preserved or will be lost, in the end. $\endgroup$
    – Janus Boffin
    Commented Apr 26, 2016 at 9:41
  • $\begingroup$ And the answer is that it doesn't matter if you take the trace or not, if you have committed yourself to an unphysical calculation, the details of the calculation don't matter. If you decide that a black hole obeys the laws of thermodynamics, i.e. it is homogeneous (enough) and in near equilibrium with its environment, then the rest is statistics. If you take the purist view that a black hole has to be considered as a fully isolated quantum system, then you have to ask yourself if a black hole can form to begin with (I do not believe it can), but you can't mix the two without getting nonsense. $\endgroup$
    – CuriousOne
    Commented Apr 26, 2016 at 9:53
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    $\begingroup$ I think this misses the question and the information loss problem. The question is about the role of partial trace, not the paradox itself. And the information loss paradox is about pure state evolving to a mixed state through unitary evolution, which cannot happen, so there is something wrong in the analysis. $\endgroup$
    – MBN
    Commented Apr 26, 2016 at 12:19
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    $\begingroup$ I do not think this answer is correct (for the reasons already given by MBN), but I appreciate you posting it properly. $\endgroup$
    – Rococo
    Commented Apr 26, 2016 at 15:23
  • $\begingroup$ @MBN: There is no paradox, that's my point. A black hole is just as much a dissipative system as a stirred cup of coffee, which is what is usually being used to highlight the problem with the Poincare recurrence theorem. Even without any loss of radiation (em, gravitational) to the outside, the smearing out due to tidal forces will make the recurrence timescale essentially infinite, much longer than the evaporation timescale. The only "information" that can effectively being retrieved would be additive scalars like mass and charge. $\endgroup$
    – CuriousOne
    Commented Apr 26, 2016 at 17:46

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