I Hamiltonian of a particle moving on the surface of a sphere

Salmone
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In a quantum mechanical exercise, I found the following Hamiltonian:

Consider a particle of spin 1 constrained to move on the surface of a sphere of radius R with Hamiltonian ##H=\frac{\omega}{\hbar}L^2##. I knew that the Hamiltonian of a particle bound to move on the surface of a sphere was ##H=\frac{L^2}{2mR^2}## and then to get the same Hamiltonian should be ##\omega=\frac{\hbar}{2mR^2}## which dimensionally fits, but is it right? How can this equality be proved?
 
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As you write yourself, it's just a definition for a parameter ##\omega##. So what should be right or wrong with it?
 
@vanhees71 So ##\omega## is just a parameter which should be equal to ##\frac{\hbar}{2mR^2}## and with dimensions ##s^{-1}##
 
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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!
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