if de Broglie's equation: wavelength = h/(mv) works allways, then what is the wavelength of an object at rest (v=0) is it undefined or infinity or something?
Because of the uncertainty principle, you can't have a wave function for a particle that is definitely exactly at rest, just as it can't have any other definite exact value of momentum (and velocity).
The best you can do for a "stationary" particle is a wave packet whose momentum has an expectation value of zero, and includes waves with both positive and negative values of momentum, in a range that is centered on zero. I suspect that such a wave packet would be a standing wave with maximum amplitude at the most likely location of the particle, and decreasing to zero in either direction so the width and the momentum range satisfy the uncertainty principle.
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles.
Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated...
Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/
by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
I don't know why the electrons in atoms are considered in the orbitals while they could be in sates which are superpositions of these orbitals? If electrons are in the superposition of these orbitals their energy expectation value is also constant, and the atom seems to be stable!