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Gravitons and decoherence

A reader of Cosmic Variance asks the following question about gravity:

But gravity works - presumably, at some level - by massive objects constantly bombarding each other with gravitons, so we are also averaging over all the possible states of gravitons that we are not keeping track of, aren't we? That should cause decoherence too, shouldn't it?
The people at Cosmic Variance seem to be rather deeply confused and impressed by this question. Because I think that there is nothing confusing or impressive about it, let me tell you the correct answer. Of course, I will agree with Sean Carroll that photons, and not gravitons, are responsible for most of the observationally relevant decoherence but there are many other issues at stake here, too.

Electromagnetism and gravity

First of all, all the qualitative facts about gravity hold for electromagnetism, too. Both forces may be visualized as an exchange of intermediate, virtual particles - photons or gravitons. And both forces may be traced to a field in spacetime - the electromagnetic field or the spacetime geometry, respectively. These fields may be excited and the excitations behave as real particles - photons and gravitons. Large, coherent configurations of these particles behave as classical waves - electromagnetic waves or gravitational waves, respectively.

But in both cases, there is a difference between many real particles and many virtual particles.

Many real particles may carry a huge entropy S. This statement means that there exist many, N distinct microstates: S is approximately equal to ln(N) times Boltzmann's constant. The entropy carried by many particle never decreases macroscopically - that's the second law of thermodynamics that defines the thermodynamical arrow of time.

Analogously, as we are forgetting some unmeasurably fine (or distant) information, we are tracing over degrees of freedom that are not interesting for us. For example, every second, a lot of new infrared photons may be emitted by the Earth: every second, we may or we have to forget about all of their velocities. By doing so, the density matrix for the remaining, measurable degrees of freedom rapidly converges to a diagonal form: that's how decoherence works.

(See pages 13-16 of this file to learn the basics of decoherence; entanglement and interpretations of quantum mechanics are discussed there, too. See also some introductory comments about quantum computing and relativistic quantum mechanics.)

The decoherence always works in the same direction of time: we may be forgetting the past but we can't ever forget the future because we have never known it, anyway. This arrow of time is called the logical arrow of time and reflects the asymmetry between assumptions and deduced predictions in Bayesian inference (or any other framework for thinking).

The processes in the previous paragraphs, namely the increase of entropy (e.g. by friction) and decoherence, are closely related and the two arrows of time, thermodynamic and logical ones, are always and inevitably aligned.

A force between two objects

Consider a proton and an electron. They attract one another, by the electrostatic force. The force may be visualized as a "constant" exchange of photons. How many photons are we talking about here? Well, this is a useful question because it is a meaningless one ;-) and the reason of its meaninglessness is illuminating.

The reason why the question is meaningless is that the photons exchanged between the proton and the electron are not "real" particles that could carry new degrees of freedom (such as the direction of motion) or information or entropy. Instead, the character and the impact of the virtual particles on the charged particles is fully determined by their sources.

The virtual particles are really just internal lines (propagators) of Feynman diagrams: in this sense there is only "one virtual photon" responsible for the whole force and its information is log(1)=0. But even if you adopt a different counting, these photons are not real particles that could carry more information if the field becomes stronger.

Because there are no new "free" degrees of freedom carried by the virtual particles, the microstates are fully determined by the states of the proton and the electron. If you quantize this system in the old-fashioned non-relativistic quantum mechanical framework, you obtain the Hilbert space of the Hydrogen atom, including the ionized states. There is a very low number of its low-lying states.

But even if you decide to use the field configurations instead of the old-fashioned force to describe the states and their dynamics, the Hilbert space will remain essentially unchanged. The only difference is that each state - for example the 1s ground state of the Hydrogen atom - will be associated with a particular pure state of the electromagnetic field, too. But these states should be identified with the states in the non-relativistic theory without fields. Quantum field theory doesn't imply any new "infinite degeneracy" of these states of charged particles.

Because we are still effectively dealing with a very small number of pure states in the Hilbert space, there is no room here for the increase of entropy and no room for tracing over inconsequential degrees of freedom. There are simply no new degrees of freedom created with time here.

So if you start with a Hydrogen atom in its ground state, it won't decohere at all. It is a complete pure quantum system described by a pure state - a localized and stable object sitting somewhere in spacetime. Pure states evolve into pure states and there is nothing to trace about - unless you want to forget about the state of the atom completely. So there is no decoherence here.

The proton and electron also attract gravitationally although the force is roughly 10^{40} times weaker than the electric force. What these two particles exchange is really a specific linear combination (one combination!) of a photon and a graviton except that the photons are much more important in this mixture. If there were a decoherence of a single Hydrogen atom caused by the virtual particles, the photons would contribute much more to it than the gravitons. But there's no such decoherence.

Two celestial bodies

You may replace the proton and the electron by two macroscopic objects - e.g. the Sun and the Earth. But everything I said is still true. The Feynman diagram would have vertices (like Sun-Sun-graviton) and the coupling constant would be greater than it was before (because the Sun is heavier than the electron) but there would still be one Feynman diagram only. The state of the gravitational field is determined by the sources. If the sources - the celestial bodies - are found in a pure state, the fields they create are in a pure state, too. And this pure state will remain a pure state forever (even though it is clearly evolving).

Even for large celestial bodies, there are no macroscopic amounts of gravitons leaving to infinity (except for a detail discussed in one minute). By this comment, we mean that the orbiting celestial bodies are not losing energy. If they're not losing energy, they can't be emitting real particles. It means that there are no new degrees of freedom created that we could trace over (or forget). There is no decoherence.

The subtlety is, of course, that in general relativity, orbiting bodies emit gravitational waves and they do lose some energy. Once again, this case is fully analogous to the case of electromagnetism. An excited Hydrogen atom emits electromagnetic waves, too (and drops to lower energetic levels). But as long as the produced waves can be described by classical configurations or coherent states, their state is pretty much unique so it doesn't carry any macroscopic entropy and there is no "macroscopic excess" of newly created degrees of freedom that could be traced over. No decoherence.

So there is a quantitative difference here. A large number of excited Hydrogen atoms emit a large number of photons. They are pretty much chaotic, carry a macroscopic entropy, and can disappear to infinity (even if they disappear inside the Sun, you may already trace over them if you believe that they will never be important again). They make both the source as well as other objects they hit in the future decohere. But as long as the waves emitted by the objects are (almost) coherent - like the gravitational waves from a binary star - there is (almost) no decoherence here just like there is almost no entropy carried away by these gravitational waves: once again, the growth of total entropy and decoherence are closely related.

By the way, if you consider a piece of dust in outer space, it decoheres very quickly. Its interactions with the cosmic microwave background are actually enough to make its position behave classically within a tiny fraction of a second: a privileged "classical" basis of localized states emerges very quickly. On the other hand, the interactions with gravitons are much weaker. So even if the number of gravitons at a certain frequency per cubed meter were equal to the number of photons - and it is much smaller in reality - they would lead to much weaker decoherence simply because their interaction with the matter is much weaker.

A break for entertainment

Sheldon didn't get the driving license at age of 16, as everyone else, because he was examining perturbative amplitudes in N=4 supersymmetric theories leading to a re-examination of the ultraviolet properties of multi-loop N=8 supergravity using modern twistor theory. Their science advisor is simply superb! There has never been a soap opera as accurate as TBBT. ;-)

Does dark matter decohere?

That's another, more sophisticated question that Sean Carroll and Daniel Holz ask. The answer is, of course, Yes. But it is indeed true that the rate of decoherence may be substantially slower than the rate of decoherence that we know from the visible world. The decoherence rate increases with
  • the number of environmental degrees of freedom that become "useless" every second
  • the accuracy with which the microstates of these environmental degrees of freedom are orthogonal to each other
  • the strength of the interactions between the environmental degrees of freedom and the relevant degrees of freedom whose states are decohering
  • the alignment between the environmental degrees of freedom and the relevant - decohering - degrees of freedom of the "measured" system
If you imagine that dark matter particles only interact through gravity, all these factors are substantially reduced. As we indicated, the number of "chaotic gravitons" that massive particles emit is substantially smaller than the number of "chaotic photons" that thermally excited atoms emit in the visible world: that's because gravity is so much weaker. And whenever you find examples where the number of real gravitons emitted is very large, they typically arise from macroscopic gravitational waves that are almost uniquely determined - so the microstates of these gravitational waves fail to be exactly orthogonal to each other and "unpredictable".

The rate of decoherence is, of course, calculable. I would indeed expect that in some appropriate way of counting, the decoherence will be 10^{40} times slower than the decoherence that fundamentally boils down to electromagnetism. But it is likely that this slowdown won't have too easily measurable (by us) consequences, anyway. If you consider much bigger pieces of (dark) matter instead of the (visible) seed of dust above and you wait a little bit longer, the decoherence eventually occurs, anyway. And it is very hard to experimentally determine whether a piece of dark matter has already decohered because it interacts with us too weakly.

In this sense, the dark matter could be in macroscopic, Schrödinger-cat-like states,
"0.6 times dead plus 0.8i times alive",
that don't decohere quickly and that would look truly bizarre in the case of visible cats. But dark matter is not so easily visible so the scales and standards of decoherence are different and much slower.

No six-feet-large Erwin Schrödinger lives in the dark matter sector who could be impressed by these macroscopic manifestations of quantum mechanics. It's no surprise: all observers who are large and long-lived enough to be intelligent are pretty much guaranteed to observe a world around them that looks classical. For intelligence, you need a large amount of bits of information to be processed and forgotten during your thinking - which is qualitatively the same condition that guarantees that decoherence occurs quickly enough.

The character of the microstates that are picked as a "privileged" basis of the Hilbert space by decoherence is analogous for dark matter and visible matter. More concretely, the microstates that decohere from each other differ by the location of some of their building blocks and the privileged states (in whose basis the density matrix is erasing the off-diagonal entries) tend to be close to eigenstates of these locations. That's because the mechanism of imprinting of the system into the forgotten environmental degrees of freedom follows the local laws of physics, much like other processes in Nature (with a possible exception of phenomena related to the event horizons and discussed at the very bottom).

So it really doesn't matter whether we consider self-interaction of dark matter or the interaction between dark matter and visible matter as causes of decoherence. Clearly, if dark matter only interacts gravitationally, its interactions with the remaining dark matter are arguably more important - because most particulate matter in the Universe is dark - and it is probably true for the rate of decoherence, too.

On the other hand, if we calculate the decoherence of ordinary visible matter, its interactions with the dark matter are negligible in determining the rate of decoherence. It's because the interaction is weak (gravity) and the dark matter is in much more coherent state, anyway.

Cosmic horizons of the unknown

So I would like to argue that all these processes that take place in limited regions of the Universe are understood, at least in principle. The rate of decoherence - i.e. the place where the classical intuition becomes a legitimate approximation of quantum mechanics - is calculable. The more "useless" and "chaotic" information about the system we study is imprinted into the environment every second, the faster the decoherence proceeds. If there are no chaotic particles escaping from the system, there is no decoherence.

It is absolutely crucial that a simple Hydrogen atom in its ground state inside an isolated box - or the vacuum itself - don't decohere. They're absolutely pure, exact, and coherent - a trivial fact that seems to be too hard to swallow for the proponents of various "atomic" theories of space which are nothing else than new models of aether. If there were an aether composed out of chaotic aether atoms (or spin networks or causal dynamical triangulations or anything else of this sort), it would induce an abrupt decoherence of everything and quantum phenomena such as interference could have never been observed.

Most of these things are thus theoretically understood. But I would like to argue that once you add event horizons or cosmological horizons, things become different. Because the "local structure" of microstates of black holes is not understood well, I believe that no one can reliably calculate the decoherence rate of a black hole. If a black hole simply emits the Hawking radiation and we want to trace over it, is there a privileged basis of the black hole microstates in which the density matrix converges to a diagonal form? How quickly does it occur? Does this question make sense? Does it have a good or mathematically elegant answer?

Needless to say, this question only becomes meaningful once you know something about the actual microstates of the black hole (which we do mostly in the case of a few classes of supersymmetric black holes). You need to know at least some basis of their Hilbert space before you discuss which basis might be picked by decoherence. ;-) It is very plausible that these privileged states might have an easy interpretation in the language of fuzzballs. However, it is also plausible that the privileged states won't be eigenvectors of "position" operators as they are in the context of non-gravitational physics. And the decoherence rate may drop close to zero, too, because the outgoing particles of Hawking radiation are effectively unable to "measure anything" about the black hole.

The context of quantum cosmology is an even tougher variation of the same question because the cosmic horizon is a more complicated version of a black hole horizon and we don't have the luxury of the asymptotic region at infinity. For example, does the thermal radiation of de Sitter space make the space (and everything inside it) decohere? How quickly does this thing occur? Does the re-absorption of the thermal particles by the horizon change the picture? I can believe that someone can give us more complete answers than those that can be generated "on the spot" but I guess that they won't be complete answers.

Unlike the case of the black holes, we don't even have any framework to answer such questions because we don't have any complete quantum theory - including the full Hilbert space - of a de Sitter space. But once again, I believe that the case of atoms and celestial bodies has been theoretically settled.

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