On Entropy Production in the Madelung Fluid and the Role of Bohm’s Potential in Classical Diffusion

Abstract

The Madelung equations map the non-relativistic time-dependent Schrödinger equation into hydrodynamic equations of a virtual fluid. While the von Neumann entropy remains constant, we demonstrate that an increase of the Shannon entropy, associated with this Madelung fluid, is proportional to the expectation value of its velocity divergence. Hence, the Shannon entropy may grow (or decrease) due to an expansion (or compression) of the Madelung fluid. These effects result from the interference between solutions of the Schrödinger equation. Growth of the Shannon entropy due to expansion is common in diffusive processes. However, in the latter the process is irreversible while the processes in the Madelung fluid are always reversible. The relations between interference, compressibility and variation of the Shannon entropy are then examined in several simple examples. Furthermore, we demonstrate that for classical diffusive processes, the “force” accelerating diffusion has the form of the positive gradient of the quantum Bohm potential. Expressing then the diffusion coefficient in terms of the Planck constant reveals the lower bound given by the Heisenberg uncertainty principle in terms of the product between the gas mean free path and the Brownian momentum.

Appendix

Assuming that the phase space density function, $\rho _{ph}(\mathbf{r},\mathbf{p},t)$, in equilibrium can be separated into coordinate and momentum parts, $\rho _{ph}(\mathbf{r},\mathbf{p},t) \equiv \rho (\mathbf{r},t)\chi (\mathbf{p},t)$, then the Boltzmann entropy becomes

$$\begin{aligned} Ent_B = -k_{B} \int \int \rho _{ph} \ln \rho _{ph} d\mathbf{r}d\mathbf{p} = Ent_S- k_{B}\int \chi \ln \chi d\mathbf{p}. \end{aligned}$$

(28)

Under the de Broglie-Bohm interpretation, the particle’s momentum is $\mathbf{p} = \nabla S(\mathbf{r},t)$, so that $\chi =\delta (\mathbf{p} -\nabla S)$. As seen from the two terms on the right-hand side of Eq. (28), the Boltzmann entropy is the sum of (i) the entropy $Ent_S$ associated with the spatial probability distribution ($\rho$) and (ii) a (formally divergent) entropy associated with the momentum probability distribution $\chi$,

$$\begin{aligned} Ent_B = Ent_S - k_{B} \ln \delta (0). \end{aligned}$$

(29)

The divergent contribution in the RHS of Eq. (29) results from the (unbounded) singular Dirac’s delta distribution for the momentum. In a discrete rendition of $\chi$ (i.e., making it a Kronecker delta), this Dirac delta distribution will be replaced by a value of unity when the condition $\mathbf{p} = \nabla S$ is satisfied, i.e., there is no entropic uncertainty associated with the momentum. Consequently, such a discrete probabilistic rendition will lead to no difference between $Ent_B$ and $Ent_S$. Thus, within the continuum framework that we work in, the singular term in Eq. (29) is an outgrowth of the fact that within the Bohmian interpretation (which indeed differs from the Madelung one [4, 7]) the variance of the continuum momentum distribution vanishes, equivalent to an assumption of zero temperature of the Madelung fluid. If, however, one interprets the de Broglie guiding equation ($\mathbf{p} = \nabla S$) only as a mean (hydrodynamic) motion, then the Madelung fluid has a non zero temperature that is proportional to the Fisher information [5], and the Boltzmann entropy in (28) will, generally, not be singular.