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Is there such a thing as self decoherence? What I Mean is, I’ve read that because macroscopic objects (like billiard bals) interact with their environment they decohere and can’t be found in a Superposition. But wouldn’t the particles interacting with themselves within the object be enough? So that even when the object would have been isolated from it’s Environment it still would decohere.

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  • $\begingroup$ There really seem to be two questions here. (1) Does "self-decoherence" prevent us from observing interference of macroscopic objects? (2) Is there some other reason why we can never observe such interference? Some of the answers seem to be not carefully distinguishing these. I think the answer to 1 is no, while the answer to 2 is yes (short wavelengths, inability to prepare the state properly). $\endgroup$ – user4552 Nov 26 '19 at 16:58
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Under (conventional) quantum theory their is no such thing as "self decoherence".

So the billiard ball, if it were fully isolated from the outside universe, would remain in a coherent (pure) state indefinitely. (In practice this could never happen. Even if placed in a perfect vacuum empty of any light the billiard ball could still decohere if it was at a finite temperature - by emitting a photon).

Some people have proposed that sufficiently heavy objects might decohere themselves through gravity, but this is a highly speculative minority opinion.


The only large object we know about that is completely isolated is the Universe itself (what else could it interact with?). In principle, the universe itself could be in a coherent state.

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  • $\begingroup$ This seems to me like the only right answer so far. The main paper where I learned about this topic was Joos and Zeh, "The emergence of classical properties through interaction with the environment," Z Phys B 59 (1985) 223. In that paper, they discuss a model involving a molecule bombarded by photons. They derive a decoherence rate $\Lambda$, which is proportional to the number density of the photons. If you set the density of photons to zero, you get no decoherence. $\endgroup$ – user4552 Nov 26 '19 at 15:05
  • $\begingroup$ If the billiard ball were warm it would radiate photons into its environment. The total state of the photons and the ball together would still be a Superposition state, but if you ignore (the technical term is "trace out") the photons then the billiard ball is in a classical state (technical term, "mixed state"). In quantum physics things act like they are classical objects whenever they are entangled with other stuff we cannot access or do not care about. This is why big things act classical, they are "too big to hide" so to speak so invariably end up entangled with something somewhere. $\endgroup$ – Dast Nov 26 '19 at 18:29
  • $\begingroup$ @BenCrowell I see you found my answer not right. I have extended it (without modifying the existing part). I would be interested to know whether, on second thoughts, you found it to be right after all. $\endgroup$ – Andrew Steane Nov 27 '19 at 15:49
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In practical terms your intuition is correct: a macroscopic object will decohere itself in the sense that it will not be possible in practice to control all the degrees of freedom sufficiently well to perform interference experiments with such an object. So in this sense Schrodinger's cat is unambiguously either alive or dead before anyone opens the box. That is, all subsequent physical phenomena are consistent with the statement "the cat is either alive, or dead, not in a superposition" and the act of opening the box is entirely irrelevant.

But this area of quantum theory, called the measurement problem, is concerned more with what may be the case in principle---might it be possible, in principle, that the patterns of nature (called laws of physics if you like) allow that macroscopic objects show interference effects? The answer to this question will depend on what is the nature of a more complete theory than the one we have now. This is because when the objects in question are large enough, one will have to account for gravitational effects along with all the other effects. It is not yet clear what is the right theory or what would be the nature of the mapping between the mathematical and physical statements of such a theory.

It may be that the interpretation problem will persist. The interpretation problem is, roughly speaking, the difference between a single-world and many-worlds interpretation. Faced with the same mathematical statements, some people draw the inference that there is one world (the one we observe) and some people draw the inference that there are many worlds---not just the one we observe but also a vast and ever-doubling collection of further worlds which do not have any physical effect on the one we observe. The first of these two positions is logically allowable if and only if there are processes in the universe which never get reversed. People holding a many-worlds view tend to say that all processes are in principle reversible. They might say, for example, that even the biochemistry of a cat proceeds by interactions which could in principle be reversed so as to reverse the death of a cat. I think we do not know physics well enough to be able to claim that with any confidence.

Added not after some comments

I see that some capable physicists have expressed doubt over whether the above is corrrect. I am therefore adding some further information.

An ordinary billiard ball has a mass of about 160 grams. The gravitational interaction energy of two such balls at $10\;$cm separation is $\Delta V = 1.7 \times 10^{11}$ J. If we suppose one ball is fixed and the other is propelled on a trajectory that would result in a superposition of two different locations separated by of order 10 cm, then the gravitational interaction would develop a phase difference between those two parts which reached $\pi$ after a time $t \simeq \hbar \pi/\Delta V = 2 \times 10^{-23}$ s. Therefore for timescales of this order or larger, the gravitational part of the interaction is non-negligible. It follows that the correct physical description must be one that incorporates both quantum mechanics and gravitation. We do not yet have a theoretical framework that is sufficiently well-established to do that. So I believe the second paragraph which I wrote above is quite correct, and it is incorrect to state that our current state of knowledge of physics is sufficient to make any confident claim about the coherence or otherwise of the motion of two such billiard balls.

Some people want to treat this scenario "under conventional quantum theory" and thus provide an answer. But what quantum theory is being employed? Is it ordinary quantum physics on a smooth space-time background? But how do we know that is right? Isn't it better to be aware of the limits of our understanding, and draw them to our attention?

There is also a difficulty with assuming unitary behaviour for macroscopic objects even in the absence of gravitational effects, because it amounts to assuming a degree of precision and correctness of a theory at a level very far beyond the one at which it has ever been tested. There is a vast parameter range for spontaneous collapse models, for example, which has not yet been ruled out experimentally.

If one thinks that billiard balls behave just like single atoms or fullerene molecules, except with more energy etc., then, yes, on such a model their coherence would be preserved (while evolving very quickly). If that is all the questioner was asking, then this is the answer. But the whole point of the questions in this area of physics is to raise a question-mark over such an assumption, and to point out that we do not know it to be true.

I don't know what the original questioner was chiefly asking about, but I think an element of it may have been an intuition that macroscopic objects such as billiard balls and cats can themselves amount to physical systems whose evolution constitutes a form of "measurement", and perhaps unitary evolution is not the whole story about the physical world. Such an intuition is correctly drawing our attention to a pertinent question to which we do not, in our present state of knowledge, know the answer.

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    $\begingroup$ My understanding of the OP's question was "hypothetically, if a billiard ball were perfectly isolated from the external universe could it remain in a pure state, or would INTERNAL forces destroy its coherence". The answer to this question should surely be "It would stay in a pure state, internal forces cannot cause decoherence." I think your answer could be improved by clarifying this right at the beginning, before the discussions of practicalities and quantum interpretations. $\endgroup$ – Dast Nov 25 '19 at 17:33
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    $\begingroup$ @Dast I see what you mean, but as a practical matter one can find decoherence here by two types of calculation. Using a "trace over environment" approach, any small part of a large thing can act as "environment" for the rest. Using a "average over detection-time" approach, the coherence phases will be evolving faster than any realistic detector. $\endgroup$ – Andrew Steane Nov 25 '19 at 19:07
  • $\begingroup$ No the cat is not unambiguously dead or alive before anyone opens the box. Not according to the standard theory anyway. It does probably have a diagonal density matrix though which means it won't be doing spooky things inside the box. In other words it will be in a state that can be represented as a superposition of classical states. $\endgroup$ – user68014 Nov 26 '19 at 10:15
  • $\begingroup$ @AndrewSteane I agree completely with your answer's content. You are saying that if your super-quantum detector includes all the billiard ball's degrees of freedom except, say, the nuclear spins of its atoms then they will decohere the rest of it. I was trying to say that I think clearly answering for the idealised, hypothetical, case before moving into practicalities would have been preferable. $\endgroup$ – Dast Nov 26 '19 at 12:45
  • $\begingroup$ Thomas Banks wrote ‘a macroscopic apparatus of modest size serves as its own “environment” (....l there is no reason to suppose that a modestly macroscopic apparatus, surrounded by a huge region of vacuum, with the latter protected from external penetrating radiation by thousands of meters of lead, would behave differently over actual experimental time scales, than an identical piece of machinery in the labaratory’ pitp.phas.ubc.ca/confs/7pines2010/individualreadings/… do you agree with this? $\endgroup$ – gerrit uitdenhagen Feb 10 at 8:40
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That is a really good question, and your intuition is largely correct.

If the macroscopic object is very small, like a buckyball or a small protein molecule, and if it is chilled to almost absolute zero so that it effectively only contains enough energy to be in its lowest possible energy state, and if its spin axis is oriented in a precisely predetermined direction (and so on); and then if it is completely isolated from its environment, then it will remain in its original quantum state. That is, its quantum state will remain coherent.

BUT, if the object does not start out in its lowest possible energy state it will actually coexist in a mixture of possible quantum states. Then things get more interesting. If the molecules in a classical container full of gas start in an arbitrary mix of kinetic energies, the gas will quickly redistribute the kinetic energies to the highest-entropy mix: the "thermal equilibribium distribution".

If a complex isolated quantum object has "excess energy" - that is, if it is in an energy state higher than its lowest possible state, it will be in a mixed state. The mixed quantum state (the "wavefunction") of the object typically will change on its own. That is, the mixture of states will change, though the set of available quantum states will not change. Another way to say it is that the probability that the object will be found in a particular state if the state is measured will change. At the end point of the evolution of the object's wavefunction, it will be in the highest entropy condition very much like the molecules in a gas if we think of each molecule in the gas as being analogous to one of the possible states of the isolated object molecule. The wavefunction evolves, so the likelihood that the object will be observed to be in any particular state when measured changes over time until it reaches that highest entropy mixture of states. In other words, its mixed state will "decohere" until that highest-entropy mixed state is reached.

Note that "isolated from its environment" is very difficult to achieve and a bit messy to define. If an object changes from one energy state to another, it will emit (or absorb) EM radiation. One way to think of it is that if the object is above absolute zero, it is like a black body: it's "warm", so it emits thermal radiation with a characteristic temperature-dependent spectrum. If we consider the thermal radiation to be part of the object, then the container needs to reflect that thermal radiation back into the container with 100% efficiency. It forms a "photon gas" that builds up until it reaches equilibrium with the object, with photons being absorbed and emitted continuously but the quantity and distribution of photons of each wavelength adding up to a black-body spectrum. Its state at that point could be said to be- and remain thereafter- coherent.

If the object's emissions are not reflected back, then energy is inevitably lost, so eventually ("eventually" could mean trillions of years!) the object will cool down to its lowest possible energy state- regardless of what state it starts in. It will then stay in that state until the end of time: it will remain coherent.

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  • $\begingroup$ In your second paragraph you write "...redistribute the kinetic energies to the lowest entropy mix...", should this read highest entropy? $\endgroup$ – jacob1729 Nov 25 '19 at 13:53
  • $\begingroup$ Of course you're right-- thanks for catching it. It's now fixed. $\endgroup$ – S. McGrew Nov 25 '19 at 14:04
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    $\begingroup$ Maybe I'm not understanding you properly, but this doesn't seem quite right to me. You seem to be talking about decoherence and thermal equilibrium as if they're the same thing, but they're not. Decoherence has to be stated in terms of some preferred basis -- if you don't pick a basis, then you can't define whether the density matrix is diagonal. What normally determines this preferred basis is the nature of the interaction with the outside world, e.g., if the interaction is one that occurs at a point, then the x basis. In the absence of such an interaction, what determines the basis? $\endgroup$ – user4552 Nov 26 '19 at 15:15
  • $\begingroup$ No, thermal equilibrium and coherence aren't the same. But while a mixed state is evolving toward thermal equilibrium, which it will do if it can, it can't really be considered coherent. At least that's true if my understanding of "coherence" is correct: that a coherent quantum state is one for which measurement outcome is predictable in a fairly strong sense. E.g., set it up in an initial state, and you can be sure it's still in the same state at a later time if that time is less than the coherence time. $\endgroup$ – S. McGrew Nov 26 '19 at 23:16
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Consider double-slit interference of large molecules. The molecules are in superposition since they interfere. But we do not record broken molecules in the experiment, which means they interfere as whole molecules. Now there is no external source (environment) for decoherence, else the molecules would not interfere. So we can see that, even in the absence of an environment, a large molecule keeps its structural integrity, which seems to mean, if you see the molecule itself as a quantum system, that it decoheres by itself.

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"Dechorence" tends to involve two ideas. One is interaction with the environment and the second is the interaction making the original system a superposition of classical states, or in more technical terms, a state that can be represented by a diagonal density matrix. A complex object isolated from the environment won't interact with the environment by definition but its self interactions will tend to put it into a diagonal density matrix type of state. An example may make this clearer. We do a Schrödinger's cat experiment but we also put a partition in the box made of plywood say. The cat is in the left half. We will (with almost 100% probability) end up with a superposition of a dead and alive cat in the left half of the box. We are vanishingly unlikely (though it is still possible) to find that the cat has quantum tunnelled across the plywood barrier. This is the distinction. I refer you to Max Tegmark's excellent book The Mathematical Universe chapter 8 for a very good clear explanation of all this.

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    $\begingroup$ I don't like Max Tegmark's work. I find he likes to wow people with crazy untestable theories, for the sake of it. I don't know what quantum tunneling has to do with this. $\endgroup$ – gerrit uitdenhagen Nov 26 '19 at 11:35
  • $\begingroup$ This is my example not Tegmark's and the part of that book I'm referring to contains no crazy untestable theories (other parts do but they are clearly signposted). The reason I included tunnelling was to demonstrate that in a way things are "decohering" inside the box even if it has no interaction with the environment. $\endgroup$ – user68014 Nov 26 '19 at 12:46

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