Bose-Einstein statistics for μ>ε

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SUMMARY

The discussion focuses on the implications of assuming a chemical potential μ greater than the energy ε in Bose-Einstein statistics. It establishes that under this assumption, the Gibbs sum simplifies to Z ≈ [λ exp(-ε/τ)]^N, leading to an average particle number ⟨N⟩ equal to N. However, it highlights a critical flaw: the Bose-Einstein distribution function f(ε) becomes negative for μ > ε, which is physically untenable. The conversation also touches on the concept of Bose-Einstein condensation, where the chemical potential approaches the lowest energy state ε₀, resulting in divergent occupation of that state.

PREREQUISITES
  • Understanding of statistical mechanics and thermodynamics
  • Familiarity with Bose-Einstein statistics
  • Knowledge of Gibbs sum and partition functions
  • Concept of chemical potential in quantum systems
NEXT STEPS
  • Study the implications of chemical potential in quantum gases
  • Explore the concept of Bose-Einstein condensation in detail
  • Learn about the derivation and applications of the Gibbs sum
  • Investigate the physical significance of negative distribution functions in statistical mechanics
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Physicists, particularly those specializing in statistical mechanics, quantum mechanics, and condensed matter physics, will benefit from this discussion. It is also relevant for researchers studying quantum gases and Bose-Einstein condensation phenomena.

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TL;DR
Flaws in Bose-Einstein statistics for μ>ε
The Gibbs sum is given by

$$Z=\sum[\lambda \exp(-\varepsilon/\tau)]^N$$

where ##\lambda\equiv\exp(\mu/\tau)##. Since we are assuming ##\mu>\varepsilon##, we take only the last term of the sum because all others can be neglected.

thus

$$Z\approx[\lambda \exp(-\varepsilon/\tau)]^N$$

Now

$$\langle N\rangle =\lambda\frac{\partial}{\partial\lambda}\ln Z=\lambda\frac{\partial}{\partial\lambda}(N\ln\lambda-N\varepsilon/\tau)=N$$


But in general, we see that for any parameter ##X##

$$\langle X\rangle=\sum X_N[\lambda \exp(-\varepsilon/\tau)]^N/Z\approx X_N$$

where ##X_N## is the value at ##N\rightarrow\infty##, thus making the whole model useless.

But is this the only flaw in taking ##\mu>\varepsilon##?

That is, for the usual Bose-Einstein distribution

$$f(\varepsilon)=\frac1{\exp[(\varepsilon-\mu)/\tau]-1}$$

we get ##f(\varepsilon)<0## for ##\mu>\varepsilon##, which is physically wrong.

Are any such "physical" conditions present for the above model?
 
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For bosons the energy ##\epsilon_0## of the lowest energy state sets an upper limit for the chemical potential. In case when the chemical potential ##\mu## approaches the lowest energy level ##\epsilon_0## from below (##\mu\rightarrow\epsilon_0##), the occupation of the lowest energy level diverges; this phenomenon is called the Bose-Einstein condensation.
 
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