Entropy Contradiction for a Single Harmonic Oscillator

[Snapu]

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 σ_N. 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??

 

[Snapu]

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.

 

[kafri]

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.