# Does black hole evaporation respect quantum superpositions?

I've confused myself about the following scenario:

Suppose you make a black hole out of states with spin aligned into one direction, say the positive x-direction, and let's call this "up". Then the so-created black hole has an non-zero angular momentum because the expectation value of the spin is nonzero. (Let's assume there's no other net contribution to the angular momentum, just for simplicity. I don't think this matters.) The so-created black hole evaporates in the semi-classical limit with an energy spectrum that depends on this expectation value of the total angular momentum. (Forget about what happens in the final stage where the semi-classical approximation breaks down, I don't think this matters.)

Next you make a black hole out of the same initial state with the only difference that all spins are down. Same thing: the emission spectrum - the distribution of outgoing particles over energy - depends on the expectation value of the angular momentum, at least in the semi-classical limit we're pretty sure of that.

Finally, you make a black hole out of a superposition of the two initial states. The expectation value of the total angular momentum is now zero, correspondingly the angular momentum in the semi-classical Kerr-background is zero. What is the decay spectrum (in the early radiation) in this case?

Is the outgoing radiation a) a superposition of the radiation that one obtained from the purely up and purely down case? But then it would depend on the angular momentum which the black hole doesn't have. Or does it b) not depend on the angular momentum of the superposed states (note that what is and what isn't a superposition is ambiguous anyway) but then the evolution doesn't seem to be linear? (In the sense that |psi_in> + |psi'_in> doesn't result in |psi_out> + |psi'_out> when |psi_in> resulted in |psi_out> and same for the dashed case).

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There are indeed many confusing statements implicitly included in the OP's question.

First, the superposition principle holds everywhere in Nature. Our Universe respects the postulates of quantum mechanics and the linearity of the wave function is one of these postulates that is valid completely universally. In particular, all observables including the Hamiltonian and the unitary evolution operators have to be linear.

Second, it is not true that the superposition of two states with a nonzero angular momentum has a vanishing angular momentum. Quite on the contrary, one may show that the probability that the superposition has $J^2=0$ is equal to zero if both states $\psi_1,\psi_2$ in the superposition are eigenstates of $J^2$ with $J^2\neq 0$. What you're probably confused by is the fact that the expectation value of $J_z$ may be zero in a superposition state. But that's something completely different than saying that $J_z$ itself is equal to zero, or even $J^2=0$. For example, a superposition of electron's up- and down- state is always just another state with $j_a=\pm \hbar/2$ with respect to some axis $a$, and this state always has $J^2=j(j+1)\hbar^2$ for $j=1/2$, regardless of the coefficients.

Third, a black hole can't really be constructed out of several light enough spinning particles in the same direction because it would violate the extremality bound for the Kerr or Kerr-Newman black hole. The angular momentum would be too high and similarly for the charge if you took electrons.

Fourth, the evolution of a linear superposition is completely dictated by the evolution of the individual terms, exactly because of the linearity of quantum mechanics. So if you know what is the final state for the initial states $\psi_1,\psi_2$, you also know what's the final state for their superposition. Probabilities and expectation values in such a final state are in between the corresponding ones for $\psi_1$ and $\psi_2$ but they may also include "interference terms" mixing $\psi_1$ with $\psi_2$.

Quite generally, the points 1,2,4 indicate that you don't appreciate that the wave function is a template for the probability distributions and not an actual classical wave of any kind. And that the expectation values aren't the values that one will actually measure: they're just averages of the values one may measure.

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i had the impression that whenever angular momenta or charge was added to a black hole to produce super-extremality, the black hole mass would always increase to leave the black-hole in a extremal or sub-extremal state. Why isn't that what happens in this case? the black hole formed by all those spin-aligned electrons/positrons shouldn't have a squared mass of $M^2 + a^2$? –  lurscher Dec 24 '12 at 15:21
Yes, I am talking about the expectation value of a projection of the angular momentum on one particular axis being zero in the superposition state. Are you trying to tell me that it is not this expectation value that enters the angular momentum parameter of the Kerr-metric in the semi-classical case?? Your arguments 1,4 are totally besides the point. It is because I think this should be so that I am asking. 3, well mix in some non-spinning heavy particles then. Unless you want to claim you can't make any construction of the type I talk about it doesn't matter. –  WIMP Dec 25 '12 at 7:50
Updated the question to make it clearer, hope that helps. –  WIMP Dec 25 '12 at 10:58