Details

The three-dimensional stationary Schrödinger equation with Coulomb potential can be written in parabolic coordinates as

,

with the wavefunction , the Laplace operator , the partial derivative with respect to time and the Coulomb potential . For simplicity, we set atomic units .

In spherical polar coordinates, the unnormalized energy eigenfunctions for the hydrogen atom associated with the quantum state with the quantum numbers is given by:

,

where is a spherical harmonic. The radial function is

,

where is a generalized Laguerre polynomial.

For a stationary case with , the phase function reduces to zero. Thus the Bohm momentum of the ground state electron is , and finally the electron is at rest [1]. For , the electron is in motion.

If the wavefunction is expressed in parabolic coordinates (for more details see [4–6]), we find:

,

where

,

,

.

Here , and is a Whittaker function.

The energy spectrum depends only on the integers , and .

For the anisotropic Coulomb potential, the Schrödinger equation in Cartesian coordinates is:

with

,

and so on.

Due to computational limitations, the superposed total wavefunctions are taken as a sum of just three eigenfunctions , where the unnormalized wavefunction for a particle is written as:

,

with and (here, ).

Each part of the superposition of the three stationary state has the same energy:

The parabolic wavefunction could be expressed by a linear combination of stationary states of spherical wavefunction . For example,

.

Let for convenience. After variable transformations from parabolic coordinates to Cartesian coordinates, we obtain the total phase function :

The gradient of the phase is obtained from the total wavefunction in polar form, , with the quantum amplitude . The velocity field is calculated by the Bohm momentum with mass . From the Bohm momentum it follows that

(here, ).

In the case of the superposition of the three stationary states with equal energy, the corresponding differential equation system for the velocity field becomes autonomous and obeys the time-independent part of the continuity equation with the wave density .

For the wavefunction in parabolic coordinates reduces to ,

with the total phase function :

.

This could be simplified for :

.

The trajectories reduce to circles with the velocities:

,

and

.

The velocity depends on the position of the particle only. For and , most of the orbits form "spirally circles" along the axis, in which a spiral orbit is perpendicular on the circle orbit (for an example, see [7]).

In all other cases, the integration of the velocity vector in respect to time leads to complex periodic or chaotic orbits (for examples, see [8, 9]). If with (), it is strongly implied that chaotic motion occurs.

When PlotPoints, AccuracyGoal, PrecisionGoal and MaxSteps are increased (if enabled), the results will be more accurate.

References

[1] P. R. Holland, The Quantum Theory of Motion: An Account of the de Broglie–Bohm Causal Interpretation of Quantum Mechanics, New York: Cambridge University Press, 1993.

[2] Bohmian-Mechanics.net. (Dec 26, 2022) bohmian-mechanics.net.

[3] S. Goldstein, "Bohmian Mechanics." The Stanford Encyclopedia of Philosophy (Summer 2017 Edition). (Dec 26, 2022) plato.stanford.edu/entries/qm-bohm.

[4] L. I. Schiff, Quantum Mechanics, 2nd ed., New York: McGraw-Hill, 1955.

[5] E. Merzbacher, Quantum Mechanics, New York: Wiley, 1961.

[6] J. F. Ogilvie, "The Hydrogen Atom According to Wave Mechanics: Part II. Paraboloidal Coordinates," Revista de Ciencia y Tecnologia, 32(2), 2016 pp. 25–39. arXiv:1612.05098.

[7] K. von Bloh. Bohm Trajectories for the Anisotropic Coulomb Potential with Constant Phase Shift: Part 3 [Video]. (Dec 28, 2022) www.youtube.com/watch?v=kIxbf256IlE.

[8] K. von Bloh. Bohm Trajectories for the Anisotropic Coulomb Potential in Parabolic Coordinates: Part 1 [Video]. (Dec 26, 2022) www.youtube.com/watch?v=mhkX1VArlT0.

[9] K. von Bloh. Bohm Trajectories for the Anisotropic Coulomb Potential with Constant Phase Shift: Part 2 [Video]. (Dec 26, 2022) www.youtube.com/watch?v=mmNFFvksa8Q.