Making use of the partition function, it is straight forward to show that the entropy of a single quantum harmonic oscillator is:(adsbygoogle = window.adsbygoogle || []).push({});

$$\sigma_{1} = \frac{\hbar\omega/\tau}{\exp(\hbar\omega/\tau) - 1} - \log[1 - \exp(-\hbar\omega/\tau)]$$

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 σ_{1}.

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??

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# I Entropy Contradiction for a Single Harmonic Oscillator

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