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A classical harmonic oscillator has energy given by $\frac{1}{2m}p^2+\frac{1}{2}kx^2$. This means its Boltzmann factor is

$$e^{-\frac{\beta p^2}{2m}}e^{-\frac{\beta k x^2}{2}}$$

where $\vec{x}$ and $\vec{p}$ are the continuous position and momentum vectors, respectively. The partition function should therefore be given by

$$Z=\int e^{-\frac{\beta p^2}{2m}}d^3\vec{p}\int e^{-\frac{\beta k x^2}{2}}d^3\vec{x},$$

but it is stated in my course homework that the partition function is instead

$$Z=\frac{1}{h^3}\int e^{-\frac{\beta p^2}{2m}}d^3\vec{p}\int e^{-\frac{\beta k x^2}{2}}d^3\vec{x}.$$

Some sources online have instead a factor $\frac{1}{h}$ but without any justification. Either way, I cannot see how $h$ enters into this calculation. Where does it come from?

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  • $\begingroup$ Isn't it just some dimensional analysis ? p*x having the dimension of h. So that'd go in the denominator of your d^3 p. $\endgroup$ – picop Oct 27 '20 at 11:24
  • $\begingroup$ Sure, but there may be a factor there as well. $\endgroup$ – Pancake_Senpai Oct 27 '20 at 11:33
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Classical partition function is defined up to an arbitrary multiplicative constant. dividing it by $h$ is done traditionally for the following reasons:

  • In order to have a dimensionless partition function, which produces no ambiguities, e.g., when taking its logarithm
  • It provides a smooth junction with the quantum case, since otherwise some of the quantities would differ due to the arbitrary choice of the constant in the classical case, which is however not arbitrary in the quantum treatment.

And many textbooks do explain this.

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