Entropy Contradiction for a Single Harmonic Oscillator

  • #1
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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??
 

Answers and Replies

  • #2
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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.
 
  • #3
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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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