Entropy Contradiction for a Single Harmonic Oscillator

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SUMMARY

The discussion addresses the entropy of a single quantum harmonic oscillator, revealing a contradiction between the entropy calculated using the partition function and the multiplicity function. The entropy for a single oscillator is derived as σ₁ = (ħω/τ)/(exp(ħω/τ) - 1) - log[1 - exp(-ħω/τ)], while the multiplicity function yields an entropy of zero, indicating a discrepancy. For N oscillators, the partition function Zₙ = Z₁ⁿ confirms consistency, producing σₙ = Nσ₁. This inconsistency for N=1 highlights the unique behavior of quantum systems compared to classical ones, where the entropy is fixed at one Boltzmann constant.

PREREQUISITES
  • Understanding of quantum mechanics and harmonic oscillators
  • Familiarity with statistical mechanics concepts, including partition functions
  • Knowledge of entropy calculations in thermodynamics
  • Basic grasp of Boltzmann's constant and its significance
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  • Study the derivation of the partition function for quantum harmonic oscillators
  • Explore the relationship between entropy and temperature in statistical mechanics
  • Investigate the implications of the multiplicity function in thermodynamic systems
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Making use of the partition function, it is straight forward to show that the entropy of a single quantum harmonic oscillator is:
$$\sigma_{1} = \frac{\hbar\omega/\tau}{\exp(\hbar\omega/\tau) - 1} - \log[1 - \exp(-\hbar\omega/\tau)]$$However, if we look at the partition function for a single harmonic oscillator, then it is just g(1,n) = 1.
If we take the the logarithm of g(1,n), then we get the entropy is 0 which is in direct contradiction to our above result for σ1.

What is going on here?

This is all the more confusing because you can show that the partition function and multiplicity function approach are consistent for the N oscillator system. Namely, given that the partition function for the N-oscillator system is:
$$ Z_{N} = Z_{1}^{N}$$
We can readily get the result:
$$\sigma_{N} = N\sigma_{1}$$

We can then check this result by starting with the multiplicity function for N harmonic oscillators:
$$ g(N,n) = \frac{(N+n-1)!}{n!(N-1)!}$$
take the logarithm to get the entropy, and then use the definition for temperature
$$\frac{1}{\tau} = \frac{\partial\sigma}{\partial U}$$

Cranking through this, we can reproduce our above result for σN. Thus, for the N harmonic oscillator problem, calculating the entropy with a partition function approach is shown to be consistent with a multiplicity function approach even though it is not true for N=1. HOW??
 
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EDIT: Above I wrote:
However, if we look at the partition function for a single harmonic oscillator, then it is just g(1,n) = 1
I meant to say that the multiplicity function is g(1,n) = 1. Not the partition function.
 
The energy of classical harmonic oscillator in a bath in equilibrium is kT where k is Boltzmann constant and T is the temperature of the bath. The entropy is S=Q/T=kT/T=k.
Therefore a classical harmonic oscillator has a fix entropy of one Boltzmann constant. Quantum oscillator has lower entropy that reduces exponentially with energy.
 

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