Is Degeneracy in Quantum States Equally Probable in Thermodynamic Equilibrium?

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Discussion Overview

The discussion centers on the concept of degeneracy in quantum states and whether these states are equally probable in thermodynamic equilibrium. It explores the implications of the Boltzmann distribution and the postulate of equal a priori probability within the context of statistical mechanics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant references a Wikipedia definition of degeneracy, questioning the assertion that degenerate states are equally probable.
  • Another participant supports the idea by citing the Boltzmann distribution, suggesting that the probability density for particles in degenerate states should be the same if they share the same energy level.
  • A third participant introduces the concept of mutual accessibility and thermodynamic equilibrium, mentioning the "Postulate of Equal a Priori Probability" as relevant to the discussion.
  • Another participant agrees with the notion of equal probability in thermodynamic equilibrium, noting the tendency for particles to occupy lower energy levels over higher ones.

Areas of Agreement / Disagreement

Participants express varying degrees of agreement on the relationship between degeneracy and probability in thermodynamic equilibrium, but the discussion does not reach a consensus on the implications of these concepts.

Contextual Notes

The discussion does not address potential limitations or assumptions underlying the Boltzmann distribution or the postulate of equal a priori probability, leaving these aspects unresolved.

Who May Find This Useful

Readers interested in statistical mechanics, quantum physics, and the principles of thermodynamic equilibrium may find this discussion relevant.

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 Boltzmann 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
 

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