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Quantum decoherence is an irreversible process which is the result of interaction of the system with its environment. It prevents interference due to lack of coherence. Environment acts just like a heat bath. Now my question is, is the different branches of the wave function of the universe becoming gradually more and more decoherent so that in a far future no trace of interference can occur? In other words I would like to know whether there will be any quantum death (like heat death of thermodynamics) of the universe if one waits for a sufficiently long time?

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""Environment acts just like a heat bath"" I doubt that. A heat bath maybe enhances decoherence, but it does a lot more. And why is decoherence of "universe wave function" quantum death? –  Georg Feb 17 '11 at 16:41
Why would you think the universe will have a "wave function"? In my opinion the universe viewed quantum mechanically has a density matrix composed of all the zillion independent state functions of atoms, molecules, light, etc. It is thermodynamic really. –  anna v Feb 17 '11 at 16:50
@anna v: Of course there is a wave function of the universe as Hartle and Hawking showed us. Otherwise what do you mean by the subject called "quantum cosmology"? –  user1355 Feb 17 '11 at 17:00
@anna v: Are you not aware of this famous paper? link.aps.org/doi/10.1103/PhysRevD.28.2960 –  user1355 Feb 17 '11 at 17:02
Dear sb1, if you consider loss of coherence "quantum death", I assure you that 99.99999999999999999999999999999% of the quantum death is completed within a tiny fraction of a second - which may be much shorter than the Planck time for macroscopic objects. In principle, the coherence is always there if you could trace the environment, directly or indirectly, but it's totally inconsequential for physics as an empirical science. The decoherence occurs almost instantly. I still don't understand why you call it "death". It's just the appearance of the classical intuition from quantum mechanics. –  Luboš Motl Feb 17 '11 at 19:01
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1 Answer

This question is a bit strange, and I tend to agree with Anna that this is related to thermodynamics. The entropy involved here is an entanglement entropy. Suppose you have system A and system B which form an entanglement. The entanglement entropy of system A is $$ S_A~=~-Tr[\rho_A log\rho_A], $$ which equals the entanglement entropy of system B. If the two systems form a pure state then $S_{A+B}~=~0$. The entanglement entropy comes from the fact you have access to only one part of the density matrix $\rho~=~\rho_A\otimes\rho_B$.

This plays a role in cosmology at large. During inflation the vacuum energy density was huge. The cosmological constant $\Lambda~\simeq~(8\pi G/c^4)\rho$, is very large and drives a rapid exponential expansion of space. There is some theoretical controversy here, but while the energy density of the vacuum was very large, the entropy was not that large. The entropy is a measure of the number, N, of degrees of freedom in a system that are coarse grained into a macrostate $S~=~k log(N)$. The other oddball factor is that while the temperature was high, the entropy was low due to the negative heat capacity of event horizons in spacetime. During inflation the event horizon was smaller than a proton, and the entropy is proportional to the area of the horizon $S~=~k A/4L_p^2$.

The bang came about because the exponential expansion rapidly came to a halt, the cosmological constant dropped to a small value (the vacuum energy dropped enormously) and the cosmological horizon adjusted to a very large value. It is now out about $10^{10}$ light years. This means a relatively small number of degrees of freedom enter into complicated entanglements which are not accessible in a local region. The entanglement entropy increases, and these states appear in a highly thermalized form. This is the bang and fire of the big bang. It is a form of latent heat of fusion in a phase transition. The large vacuum energy $\rho~\simeq~10^{100}GeV^4$ crashed into about 10 GeV^4, and the energy gap assumed the form of a thermalized gas of particles.

This was the initial generation of a huge amount of entropy in the early universe. Subsequently entropy is in the form of black holes, radiation and so forth. It is interesting to think we can understand this all from the perspective of quantum mechanics. Into the future the universe will end up as a de Sitter vacuum. In the question:

de sitter cosmologic limit

I indicated how the universe will over a vast period of time will decay from the de Sitter vacuum configuration with a small vacuum energy to a Minkowski spacetime. The horizon will retreat of to “infinity,” which means the entropy becomes infinite. It might be problematic to think of infinite entropy. There might be some sort of cut-off in this process. On the other hand this is just a measure of how the vacuum decays away to zero and there is no energy. The retreat of the cosmological horizon off to infinity is probably a measure of continued quantum entanglement process, which proceeds almost indefnately.

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Thanks for the answer. Of course it is related with thermodynamics and entropy generation but what I really wanted to know is whether it (decoherence) means a complete absence of any interference in the long run. –  user1355 Feb 17 '11 at 17:24
In a measurement of a superposed pair of states the superposition or overlap is replaced by an entanglement. So what you ask could be answered in the affirmative. –  Lawrence B. Crowell Feb 17 '11 at 19:09
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