Why Normalise by $h$ in the Partition Function for Classical Harmonic Oscillator?

Question

I was wondering if anyone could explain the reasoning behind the $h$ normalization constant when calculating the partition function for a classical harmonic oscillator.

I know that the partition function is $Z=\sum e^{-\beta H(\vec{x},\vec{p})}$ where $H=\frac{p^2}{2m}+\frac{kx^2}{2}$. I also know that to compute this we need to do an integral instead of a sum (as $\vec{p}$ and $\vec{x}$ are (classically) continuous).

So I understand why we go from $Z=\sum e^{-\beta H(\vec{x},\vec{p})}$ to $$Z=\frac{1}{N^3}\int d^3p \int d^3x e^{-\beta H(\vec{x},\vec{p})}$$

where $[N] \sim \mbox{kg} ~ \mbox{m}^2 ~ \mbox{s}^{-1} $ for dimensions.

I just don't understand how we go from this to concluding that $N=h$.

I understand it has something to do with $dx ~ dp \sim h$, but I'm finding it difficult to motivate/justify this from a non-quantum perspective, seeing as this is a classical oscillator. Is there any way, or does this just have to be motivated from a quantum perspective?

Answer 1

The $N$ is not really the Planck's constant $h$. It is denoted as such because that was the convention. This has to do with the history of the subject.

Statistical mechanics, in its classical form was developed much earlier and as a result this equation was already known before Planck established the Planck's constant. Now, even in classical mechanics, the phase space volume must be taken to be 'something'. As a result, it was sometimes denoted by $h$. Now, when Planck solved the problem of black body radiation, this constant obviously arrived there as well. Remember he used the semi-classical approach of treating photons as oscillators in a cavity, the exact equation of which you have given here (which was later rectified by Bose-Einstein in their quantum statistics). So, this constant some guess that Planck gave it the name the hypothesis constant and hence $h$. Although this is debated.

But now that quantum statistical mechanics is well known, anticipating that the smallest phase space cell will be of the order of $h$, it is such equated to it. Otherwise, in classical stat mech, its merely an arbitrary constant with not much of a physical significance apart from normalization.

Answer 2

In classical continuous statistical mechanics, we still introduce a notion of a microstate. The following answer is extracted from Wikipedia:

A microstate occupies an extended region in the $2n$-dimensional phase space that has a particular volume $h^n$.

This equal-volume partitioning is a consequence of Liouville's theorem, i.e., the principle of conservation of extension in canonical phase space for Hamiltonian mechanics.

Here $h$ is an arbitrary but predetermined constant with the units of energy$\times$ time, setting the extent of the microstate.

Since $h$ can be chosen arbitrarily, the notional size of a microstate is also arbitrary. Still, the value of $h$ influences the offsets of quantities such as entropy and chemical potential, and so it is important to be consistent with the value of $h$ when comparing different systems.

Since the advent of quantum mechanics, $h$ is often taken to be equal to Planck's constant in order to obtain a semiclassical correspondence with quantum mechanics.