Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The concept of information loss is usually discussed with respect to a black hole. My understanding is that whatever matter you put into the black hole, it has only 3 "hairs" and so one doesn't know, just by determining the properties of the black hole, the mechanism by which the black hole was formed. There have been many developments with many people now believing that information is not really lost but gets mangled, etc.

Why is this loss of information not discussed in a far more pedestrian context? If you have a particle and an anti-particle annihilating into two photons, say; by observing the photons, you cannot reconstruct the velocities of the two particles. Have you lost information in this case? Is this concept of information identical to the Shannon definition? If annihilation is unitary, with entropy being conserved, I understand that Shannon information is also conserved. But, we cannot reverse-evolve to a unique initial state, can we? (Velocity isn't Lorentz-invariant, but, let us say that everything is carried out in a single inertial frame.)

More generally, I don't understand how information is not lost in so many processes that are many-to-one because of the nature of particle physics and why this is different from the scenario with black holes.

share|improve this question

2 Answers 2

QFT is time reversible and unitary. That the outcome of a particular scattering experiment (in this case $e^{+} + e^{-}\rightarrow 2\gamma$) is random doesn't mean that you couldn't construct the initial state assuming that you reproduced the experiment many times, and did nothing but measure the final photon states.

It's the same thing as saying that you don't know where a particular electron in the double slit experiment will end up, but you will know the final distribution of many similarly prepared electrons entering a double slit apparatus.

In this link the cross-section for this collision is found in terms of the Mandelstam variables commonly used to encode the momenta of particles in particle physics. Note that the answer depends on $s,t,$ and $u$. Since the differential cross section is measurable, this means that this experiment lets us measure our mandelstam variables, and therefore, we can determine information about the momentum of the electron, independently of the momentum of the positron.

share|improve this answer

I am going to venture and propose a possible solution through the Church of the larger Hilbert space. Before being radiated from the blackhole, the particle interacts with other particles at or near the horizon and becomes entangled with them. If so, no information is actually lost, the state became mixed only because it became entangled. If the entire black hole were to evaporate hypothetically, all information would be restored to the universe.

(As an aside, here is a mechanism how this could happen. We have our particle in state $\rho_1$ and another particle in state $\rho_2$. They interact with some unitary $U$, so that we have $U (\rho_1 \otimes \rho_2) U^\dagger$. Subsequently, particle one escapes from the black hole, while the particle 2 stays inside. Upon exiting, the first particle's state is $F(\rho_1) = \text{Tr}_2[U (\rho_1 \otimes \rho_2) U^\dagger]$. This is very much like decoherence and it is usually (but not necessarily) the case that $S(F(\rho_1)) \geq S(\rho_1)$. The kind of operations where entropy is increased are common when interacting with the environment and it is precisely entangling operations with the environment - also called decoherence - that cause entropy increase in states and the emergence of the classical world. For a more detailed read on this, see this execellent review by Zurek: Decoherence, einselection and the quantum origins of the classical)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.