How to think about speed or velocity of an electron (in an atom)?

Question

One answer to this question explains that "velocity of electrons has no meaning" while another says that "it can be argued that they don't move around atoms at all". And then in another post it is answered that electrons in atoms do indeed have a speed associated with their kinetic energy. Granted, speed is not velocity, but is the expectation value $\langle\hat{p}^2/m^2\rangle$ not what one can think of as akin to a classical "speed" (squared). Is a measured value of $\hat{p}/m$ not akin to a classical velocity* (or can this not be measured, not even a probability distribution)?

How is (a chemist) to think of angular momentum of an electron in an atom (specifically orbital angular momentum)? How completely should one abandon the classical idea of velocity as part of momentum when discussing quantum particles? This post has an answer that provides a geometric argument about "transforms under rotations". Isn't there a more (classically) appealing picture?

$*$ ignoring relativity

Answer 1

Velocity is an observable in quantum mechanics. For a massive particle, the velocity operator is the momentum operator divided by the mass. Eigenstates of energy in an atom are not eigenstates of velocity. If you prepare a hydrogen atom in a state of definite energy, then measure the velocity of the electron, you get a random vector that has a certain probability distribution. The mean is zero.

I don't think any of this really contradicts the statements you list from the other places on this site. Those statements may appear to contradict each other, but that's because they're using imprecise language. When you use English to talk about quantum mechanics, it doesn't necessarily fit.

Answer 2

In chemistry, angular momentum is often zero.

The situation in molecules or solids is quite different from free atoms. Angular momentum is a good quantum number for free atoms because of spherical symmetry. The spherical harmonic functions $Y_{\ell,m}$ are solutions of the angular part of the Schrödinger equation. These are the traveling waves where the phase of the wave function propagates around the nucleus.

In molecules, there is no spherical symmetry. The wave functions are often standing waves, where for example a $p_x$ orbital is a sum of counterpropagating $Y_{1,1}$ and $Y_{1,-1}$. One cannot see anything rotating. In the language of solid state physics of magnetism of the $3d$ transition metals one says that the angular momentum is quenched.

Answer 3

I want to talk about the well-known analogy between the Schrödinger equation and the Navier-Stokes equation. It reveals the connection between the solution of the Schrödinger equation and Navier-Stokes. In this case, the Schrödinger equation describes the continuity equation. It is similar to the continuity equation with a density equal to $\rho=\psi^*\psi$. We write the Schrödinger equation and transform it using the identity $\frac{\partial^2 \psi}{\partial x_l^2}=\psi[\frac{\partial^2 ln \psi}{\partial x_l^2}+\frac{1}{\psi^2}(\frac{\partial \psi}{\partial x_l})^2]$ $$i\hbar\frac{\partial \psi}{\partial t}=\frac{-\hbar^2}{2m}\sum _{l=1}^3 \frac{\partial^2 \psi}{\partial x_l^2}+U\psi=\frac{-\hbar^2}{2m}\psi[\frac{\partial^2 ln \psi}{\partial x_l^2}+\frac{1}{\psi^2}(\frac{\partial \psi}{\partial x_l})^2] +U\psi$$

Dividing by mass $ m \psi $, we obtain the equation $$\frac{i\hbar}{m}\frac{\partial ln \psi}{\partial t}+\frac{\hbar^2}{2m^2}(\frac{\partial ln \psi}{\partial x_l})^2=\frac{-\hbar^2}{2m^2}\sum_{l=1}^3 \frac{\partial^2 ln \psi}{\partial x_l^2} +U/m$$

We obtain the partial differential equation, taking the gradient from both sides of the equation, we introduce the real speed by the formula $$\vec V=-i\frac{\hbar}{m} \nabla ln\psi(1) $$. $$\frac{\partial \frac{ i\hbar }{m} \nabla ln \psi}{\partial t}+\frac{\hbar^2}{m^2}\frac{\partial ln \psi}{\partial x_l}\frac{\partial \nabla ln \psi}{\partial x_l}=\frac{i\hbar}{2m}\sum_{l=1}^3 \frac{\partial^2 i\frac{\hbar}{m} \nabla ln \psi}{\partial x_l^2} +\nabla U/m$$ Substituting the speed value into the transformed Schrödinger equation, we get$$\frac{\partial V_p}{\partial t}+\sum_{l=1}^3 V_l \frac{\partial V_p}{\partial x_l}=\nu \sum_{l=1}^3 \frac{\partial^2 V_p}{\partial x_l^2}-\frac{\partial U}{\partial x^p}/m,\nu=\frac{i\hbar}{2m}$$ We obtain the three-dimensional Navier - Stokes equation with pressure corresponding to the potential. The relationship between the two solutions allows us to conclude that in the microworld an imaginary solution is possible, which is obtained by substituting the real wave function in equation (1).

The imaginary kinematic viscosity of the vacuum is also awaiting an explanation. I have already raised this question in the topic "How to explain imaginary kinematic viscosity of a vacuum?", But not sufficiently argued. This topic can lead to the description of new properties of vacuum.