If I take an atom in a momentum eigenstate, or just a very narrow gaussian in momentum space (with a very large spread in position space), and then I throw it into a gas, it will quickly decohere into a much narrower gaussian in position space, and it will spread in momentum space. Is there a way to calculate the new size of the wave packet? More generally, how do I know in what way the environment will decohere my quantum state? I.e. what is the basis that my environment "measures" in?

  • $\begingroup$ Answer to part of your question: the environment measures the subsystem in whatever basis diagonalizes the interaction Hamiltonian. $\endgroup$ – DanielSank Apr 1 at 6:39

When quantitative results are desired, this kind of thing is usually analyzed in the Schrödinger picture by taking a "partial trace" over the gas degrees of freedom and considering the resulting density matrix for only the atom of interest. In that approach, a fixed basis in which the density matrix becomes and remains approximately diagonal is the basis of states selected by decoherence. This basis is slightly "fuzzy" (not quite uniquely defined) because the gas acts as an imperfect location-measuring device, but that slight ambiguity in the basis doesn't significantly affect the estimated width of the wave packet.

The caveat "remains approximately diagonal" is essential. Finding a basis that diagonalizes the density matrix exactly at just one instant in time is obviously not sufficient, because such a basis always exists at any given time, whether or not any decoherence has occurred. A basis in which the density matrix remains (approximately) diagonal doesn't always exist. When it does, that's a signature of decoherence.

Published analyses using this approach include

(Tegmark's paper is actually critiquing a modification of quantum theory, but calculations using ordinary quantum theory are included for comparison.) I don't have a copy of Joos and Zeh in front of me, but Tegmark at least provides a quantitative estimates of the size of a particle's wavepacket after decoherence is well-developed, as well as the time required for decoherence to become well-developed. The width is $$ \Delta x \sim\frac{\hbar\tau}{m\lambda_\text{eff}} $$ where $m$ is the mass of the atom, $\lambda_\text{eff}$ is an effective wavelength of the gas particles, and $\tau$ is the coherence time, which in turn is given by $\tau=1/(\sigma\Phi)$ where $\sigma$ is the total scattering cross section and $\Phi$ is the average flux of gas-particles per unit area per unit time (regarded as a flux of incident particles being scattered by the atom of interest).

In principle, when the whole system is considered, so that the state-vector includes all of the details of the gas as well as the atom, the states selected by decoherence are (by definition) the states that cannot be mixed with each other by the operators representing any feasibly-measurable future observables. (The word "future" is redundant in the Schrödinger picture, but I included it so that the statement is also valid in the Heisenberg picture.) The density-matrix approach cited above is consistent with this. Again, the delineation between these states is slightly "fuzzy" because the gas acts as an imperfect location-measuring device, but the basic principle still applies.


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