The energy-speed relationship of quantum particles challenges Bohmian mechanics?

TL;DR

The paper addresses whether a recent experiment claiming a violation of Bohmian mechanics' phase-speed relation is valid. It argues the experiment relies on a simplified 1D model and demonstrates that a two-dimensional Schrödinger treatment with a separable Hamiltonian ${\hat{H}}={\hat{H}}_x+{\hat{H}}_y$ yields a nonzero phase gradient $\vec{\nabla}S$ from the y-dynamics, compatible with the observed density current. The key finding is that the x-direction density transfer is governed by transverse (y) oscillations between even and odd y-states, which preserve Bohmian dynamics while explaining the measured signals. This work reinforces Bohmian dynamics against the claimed violation and suggests a robust bidimensional framework for interpreting tunneling in waveguide-like setups, potentially establishing new standards for tunneling measurements.

Abstract

Recently, Sharoglazova et al. claimed to have proven a violation of the basic tenet of Bohmian mechanics, namely the phase-speed relation $\vec{v}(\vec{r},t)=\frac{\hbar}{m}\vec{\nabla}S(\vec{r},t)$. Here, $S(\vec{r},t)$ is the (real) phase of the wave function $ψ(\vec{r},t)=ρ^{\frac{1}{2}}(\vec{r},t)e^{iS(\vec{r},t)}$. In a nutshell, they have measured the speed of a claimed evanescent wave, which is real and therefore must have $\vec{\nabla}S=\vec{0}$. However, Fig. 2 clearly shows a density motion from one waveguide to the other, implying a nonzero density current, $\vec{j}(\vec{r},t)=\frac{\hbar}{2mi}\Im(ψ^*\vec{\nabla}ψ)$. If we combine this evidence with the mathematical identity $\vec{\nabla}S=\frac{m}ρ\vec{j}$, we should instead conclude that $\vec{\nabla}S\neq\vec{0}$. So, where does this apparent inconsistency come from?

Figures (1)

• Figure 1: Here $\frac{\partial S(y,t)}{\partial y}$ is represented (in $\mu$m$^{-1}$), as a function of $y$ (the single harmonic-oscillator minima are at $y_{\rm min}^{\pm}=\pm 10$$\mu$m) and $t$ (in the range $[0,T]$, with the period $T\simeq 80$ ps). At $t=0$ the current is negative, because the spectral density moves from the positive to the negative minimum.