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Is there an ergodicity assumption in quantum statistical mechanics ?

The classical statistical mechanics derives its main results from the assumption that all the states with the same energy (and particle number) are equally probable. The assumption is often formulated as the ergodicity hypothesis (perhaps postulate is more precise a word here): since the energy is conserved, the trajectory of the system in a phase space will pass over time through all the states with the same energy, hence the time average can be replaced by the ensemble average.

The equivalent results of the quantum statistical mechanics are usually derived by assiming that we cannot discriminate between the energy states in a small energy interval, $\Delta E$ (or a small volume of the phase space, $\Delta \Gamma$), and therefore can treat them as equally probable. This assumption is thus grounded in the limitations imposed by the Heisenberg uncertainty relation and the precision of our measurement device - we cannot distinguish energies closer than $\sim 1/\Delta t$, where $\Delta t$ is the characteristic time of our measurement device. Moreover, in some cases the time scale is set by even more fundamental factors - e.g., $\Delta t$ cannot be longer than the lifetime of the universe.

Does this mean that the ergodicity hypothesis is redundant? Does it mean that the necessity for statistical mechanics description is dictated by the quantum mechanical nature of the universe (rather than the computational convenience of reducing multiparticle system to a few essential parameters)?

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  • $\begingroup$ This assumption is thus grounded in the limitations imposed by the Heisenberg uncertainty relation and the precision of our measurement device<<< I cannot agree on this statement. If the uncertainty principle is related to the ergodic assumption of statistics, then there would be no uncertainty in energy, since all of the visited or unvisited states are of the same energy. $\endgroup$
    – ytlu
    Jul 17, 2021 at 13:33
  • $\begingroup$ @ytlu it is not quite clear to me, which is why I ask this question. You are welcome to write an answer, explaining how the ergodicity enters the quantum statistical mechanics. $\endgroup$ Jul 17, 2021 at 13:38
  • $\begingroup$ The uncertainty of quantum is a pure math relation between a pair of variables conjugated by the Fourier transformation, $k$ and $x$, $\omega$ and $t$. It had been observed in wave equation, before quantum mechanics (but of course not the term.) The ergodicity is also taken by granted in quantum statistics, to assume all number of microscopic configurations renders the entropy of thermal dynamics. These two concepts are different and unrelated in my percept. $\endgroup$
    – ytlu
    Jul 17, 2021 at 13:51
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    $\begingroup$ I'm not sure this is completely settled yet, but research related to the Eigenstate Thermalization Hypothesis might be helpful. One of the earlier papers is Quantum mechanical evolution towards thermal equilibrium. $\endgroup$ Jul 17, 2021 at 13:53

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There are several issues involved in the post. Some are relevant, some are not. I would like first to clarify three different concepts mentioned here.

  1. Uncertainty principle: it is a math relation between two conjugate variables. The two conjugate variables are linked by the Fourier transformation. As long as a pair of variables is conjugated by Fourier transformation, the variances of their distribution functions is inversely related. For the people tailoring shape of wave train, it is a common sense that sharper a wave profile, broader bandwidth required. In Quantum mechanics:

\begin{align*} f(x) &= \int_{-\infty}^\infty g(k) e^{ikx} dk\\ g(k) &= \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx \\ \left(\Delta x\right)^2 &= \int_{-\infty}^\infty f^\dagger(x) f(x) (x - \bar x)^2 dx\\ \left(\Delta k\right)^2 &= \int_{-\infty}^\infty g^\dagger(k) g(k) (k - \bar k)^2 dk\\ \Delta x \Delta k &\ge \frac{1}{2} \end{align*}

  1. Ergodicity in quantum statistics. In quantum mechanics, the ergodicity is adopted as in the classical statistics without proof. Therefore, all microscopic configurations satisfied the given $N$, $V$, $E$ are taken into account, $W$, for compute entropy $S$ through Boltzmann hypothesis: $$ S = K \ln W. $$

  2. Integral infinitesimal element $\Delta E$ for the density of states. This is a math device to convert from summation over discretized energy levels into a continuum integral. It is only a math convenience, and is certainly not a necessary device. One can stay in handling the discrete level, if you are concerned about the inaccuracy of the integral. This should not be considered as an intrinsic physical property of a system.

Quantum mechanics has a built-in statistical effect, e.g. probability distribution (square of wave function), expectation value. This is a intrinsic statistical effect for those tiny particles which we are not able to trace the motion of each single one. But this intrinsic distribution is for an isolated quantum system. It has no role for temperature. When comes to the quantum statistical, the back ground picture is the quantum system brought in contact with a heat bath, which defines the temperature of the combined system. The temperature provide an extra weighting to each energy level.

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