Is Degeneracy in Quantum States Equally Probable in Thermodynamic Equilibrium?

hokhani
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In Wikipedia this sentence is written about degeneracy;
In physics, two or more different quantum states are said to be degenerate if they are all at the same energy level. Statistically this means that they are all equally probable of being filled,
do you agree with the bold statement?
 
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I think this could be correct since the Boltzman distribution is given by:

$$p(E_i)=\frac{1}{Z}\exp^{-E_i/kT}$$

so the probability density of finding a particle in two state with the same energy should be the same.
 
If they are mutually accessible, and the system is in thermodynamic equilibrium. This is the "Postulate of Equal a Priori Probability."
 
To me this makes complete sense if you consider that the system is in thermodynamic equilibrium. Think of how it is less likely that particles occupy higher energy levels than lower ones. In
 
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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