Picture a planet wandering intergalactic space. Such a planet would only couple to vacuum flucuations and the cosmic microwave background. (Ignore stray Hydrogen atoms.)

If this planet started as a pure quantum state, how fast would that state lose its coherence?

In such a system, clearly there are many more degrees of freedom that are isolated from the environment compared with those coupled to the outside. So I want to know if those isolated DOF somehow protect the purity of the quantum state.

  • $\begingroup$ I'm fairly certain through the correspondence principle that this object is well within classical realm of physics instead of quantum $\endgroup$ – Triatticus Nov 21 '18 at 19:39
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    $\begingroup$ You mean that it looses it coherence instantly? I'll point out that there's a body of literature pushing the correspondence principle further and further into the classical regime. See here nature.com/articles/nphys2863 for example. This means that in contrived and extreme circumstances (like the question above) perhaps there's more to the story. $\endgroup$ – psitae Nov 21 '18 at 19:48
  • $\begingroup$ Are you talking about a planet, or a notional special coherent condensate as heavy and big as a planet, protected against CMB interactions, somehow? $\endgroup$ – Cosmas Zachos Nov 21 '18 at 20:32
  • $\begingroup$ @CosmasZachos respond to whatever you find more interesting. You can allow or disallow CMB interactions. My questions allows them. Maybe it should disallow them, is that what you're saying? Further, beyond CMB interactions, what distinguishes a planet (starting in a pure state) from the condenstate you're talking about? $\endgroup$ – psitae Nov 21 '18 at 20:48
  • $\begingroup$ Maintaining coherence for a truly, inarguably, macroscopic object for any length of time is so far unachieved... but the ultimate goal of realistic quantum computers... their present smallness is the upper limit of the state of the art. I hope you are not asking for a tutorial on coherence times... $\endgroup$ – Cosmas Zachos Nov 21 '18 at 21:58

I am just going to quote Schlosshauer as being pertinent to this question and discussion in comments.

Reference: Decoherence and the Quantum-to-Classical transition (page 84):

To summarize, we have distinguished three different cases for the type of preferred pointer states emerging from interactions with the environment:

  1. The quantum-measurement limit. When the evolution of the system is dominated by $H_{int}$, i.e. by the interaction with the environment, the preferred states will be eigenstates of $H_{int}$ (and thus often eigenstates of position).
  2. The quantum limit of decoherence. When the environment is slow and the self-Hamiltonian $H_S$ dominates the evolution of the system, a case frequently encountered in the microscopic domain, the preferred states will be energy eigenstates, i.e., eigenstates of $H_S$
  3. The intermediary regime. When the evolution of the system is governed by $H_{int}$ and $H_S$ in roughly equal strengths, the resulting preferred states will represent a compromise between the first two cases. For instance in quantum Brownian motion the interaction Hamiltonian $H_{int}$ describes monitoring of the position of the system. However, through the intrinsic dynamics induced by $H_S$ this monitoring also leads to indirect decoherence in momentum. This combined influence of $H_{int}$ and $H_S$ results in the emergence of preferred states localized in phase space, i.e. in both position and momentum.
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    $\begingroup$ I'm glad you posted this. It makes an important point, namely that the states selected by decoherence depend on the circumstances. Another thought: Both of our "answers" cite sources that use a separation of the system into two parts, whose details we do and don't (respectively) care about, and we typically take a partial trace to hide the latter. I view this as a simplified version of what we'd really like to do, which is to separate the set of observables into those that are and are not (respectively) feasibly measureable, with only a fuzzy line between them. But that complicates the math. $\endgroup$ – Chiral Anomaly Dec 12 '18 at 3:57
  • $\begingroup$ I need to do more reading. Thanks for the textbook recommendation. $\endgroup$ – psitae Dec 13 '18 at 4:56

If this planet started as a pure quantum state, how fast would that state lose its coherence?

In this answer, I'm interpreting "started as a pure quantum state" to mean "started in a quantum superposition of different locations".

The speed of decoherence for objects of various sizes in various environments was analyzed by Tegmark in "Apparent wave function collapse caused by scattering", https://arxiv.org/abs/gr-qc/9310032. One of the environments considered in that paper is the Cosmic Microwave Background (CMB), and the largest object considered in that paper is a bowling ball. Here, I'll apply Tegmark's equation to an earth-sized planet, assuming that the only source of decoherence is the CMB.

This answer uses some numbers from Tables 1 and 2 in that paper, along with an equation from section 4.1 in that paper.

Tegmark starts with a quantum model of ordinary scattering processes, such as the scattering of CMB from the object, and derives this equation: $$ \rho(\mathbf{x},\mathbf{y},t)\approx \rho(\mathbf{x},\mathbf{y},0) \exp\left(-\Lambda t\left[ 1-e^{-r^2/2\lambda^2} \right]\right) \tag{1} $$ with $r\equiv |\mathbf{x}-\mathbf{y}|$. The quantity $\rho(\mathbf{x},\mathbf{y},t)$ is called the reduced state or the reduced density matrix. (In this case, the "matrix" has continuous "indices" $\mathbf{x}$ and $\mathbf{y}$.) It represents the degree to which a quantum superposition of different object-locations $\mathbf{x}$ and $\mathbf{y}$ remains coherent in the presence of environmental effects. If $\rho(\mathbf{x},\mathbf{y},t)$ is zero for $|\mathbf{x}- \mathbf{y}|>R$, then coherence over distances $> R$ is lost. Equation (1) tells us how quickly this happens.

Equation (1) involves two constants, $\Lambda$ and $\lambda$, with the following meanings:

  • $\Lambda=\sigma\Phi$, where $\sigma$ is the total scattering cross-section (of the bowling ball or the earth) and, in Tegmark's words, $\Phi$ is the "average particle flux per unit area per unit time." In the present example, the "particle flux" is the flux of the CMB.

  • $\lambda$ is a length scale that characterizes the effect of the environment in equation (1). According to Tegmark's Table 1, the CMB is characterized by $\lambda\sim$ 2 millimeters.

According to Tegmark's Table 2, a bowling ball has $\Lambda\sim 4\times 10^{15}\text{ s}^{-1}$. To scale this up to the earth, we can assume that the scattering cross-section is proportional to the object's cross-sectional area. Using $\sim 10$ cm for the radius of a bowling ball and $\sim 6,000$ km for the radius of the earth, the ratio of their areas is $\sim 4\times 10^{15}$. Therefore, for the earth, we have $\Lambda\sim 10^{31}\text{ s}^{-1}$. We can use this value of $\Lambda$, together with $\lambda\sim 2$ millimeters for the CMB, to draw the following conclusions from equation (1):

  • If the earth starts in a quantum superposition of two different locations separated from each other by much more than $2$ millimeters, then it decoheres (effectively "collapses" into just one of the locations) in $\sim 10^{-31}$ seconds.

  • For smaller separations, decoherence is slower, but still very fast. If the earth starts in a quantum superposition of two different locations separated from each other by roughly the width of a single atom, then it decoheres in $\sim 10^{-15}$ seconds — roughly one hundreth of the time required for light to cross the diameter of a typical hair folicle.


If this planet started as a [quantum superposition of different locations], how fast would that state lose its coherence?

Answer: Very, very quickly.

In such a system, clearly there are many more degrees of freedom that are isolated from the environment compared with those coupled to the outside. So I want to know if those isolated DOF somehow protect the purity of the quantum state.

Equation (1) depends only on the size of the object (more accurately, on its scattering cross-section), not on its internal composition. Since the earth is bound together as a single object, decoherence caused by scattering of CMB at the surface causes decoherence of the whole thing. You can't hide from decoherence by tunneling underground.

By the way, a planet-sized object always has a significant gravitational field, which influences the motion of all interplanetary molecules that are anywhere nearby. Since that influence depends on the planet's location, this is presumably sufficient to cause rapid decoherence even if the CMB were absent. But since we don't yet have an empirically-established theory of quantum gravity, I'll leave that fun thought on the shelf for now.


As pointed out in comments, my answer above only considered one example, aptly described in a comment as $|\text{here}\rangle+|\text{there}\rangle$. My answer was also specialized in another respect: it only considered decoherence resulting from the planet's interaction with external entities. That specialization is relatively easy to handle mathematically, because it ignores the object's internal structure. But a real planet has a rich internal structure on the molecular level, and the molecular complexity of the planet itself is more than sufficient to account for decoherence of its various parts. This implies, once again, that the planet's interior is not protected from decoherence, and this assertion is not restricted to examples of the form $|\text{here}\rangle+|\text{there}\rangle$.

  • $\begingroup$ Thanks Dan for a well-sources answer. It's certainly a good place to start understanding these types of phenomena. But it's important to note that pure quantum states need not be superpositions. |here> + |there> is indeed a pure quantum state, but so is |here>. So I'm reluctant to accept this as the answer. $\endgroup$ – psitae Nov 22 '18 at 23:38
  • $\begingroup$ @psitae Could you elaborate on your concern. If we take position as the decoherence basis (which is a natural thing to do here) then |here> is already decohered. $\endgroup$ – isometry Dec 10 '18 at 4:51
  • $\begingroup$ Oh, that's a good point. Actually, I'm more concerned with the energy basis. So some states at an energy |n> greater than |ground state> would decohere, right? Does it follow a similar law as the spacial superpositions discussed above? Thanks. $\endgroup$ – psitae Dec 11 '18 at 2:51
  • $\begingroup$ @psitae The general idea is that whenever we end up with a superpos'n of terms that cannot be mixed with each other by any feasibly-measurable observable, present or future, then we might as well treat those terms just like we would treat the eigenspaces of any already-measured observable. In other words, we might as well project the state onto one of those terms. This rule seems to always give the right answer, including selecting energy eigenstates of molecular electrons in the appropriate conditions. Doesn't solve the "measurement problem," though: we still can't derive Born's rule. $\endgroup$ – Chiral Anomaly Dec 11 '18 at 15:08
  • $\begingroup$ @psitae By the way, I don't think your comment will show up in Bruce Greetham's inbox unless you include the text "@BruceGreetham" somewhere in your comment. I'm not sure about that, but it might be worth checking. $\endgroup$ – Chiral Anomaly Dec 11 '18 at 15:10

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