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?

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    $\begingroup$ In classical stat. mech., the constant cannot be determined. You choose it to agree with the quantum results (which should recover the classical ones in the classical limit; whatever this then exactly means). $\endgroup$ Apr 22 at 10:58
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    $\begingroup$ "I'm finding it difficult to motivate/justify this from a non-quantum perspective" There is no way around it. The purely classical description of the universe is broken. Only a semi-classical approximation may hopefully exist. $\endgroup$ Apr 22 at 11:02
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    $\begingroup$ Yes, I agree with the previous two replies here. Phase space area $dx dp$ has a fundamental lower limit provided by quantum mechanics, so to "count states" in the partition function we add up the number of cells (weighted by the Boltzmann factor) that each have area $h$. $\endgroup$ Apr 22 at 13:38
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    $\begingroup$ You might wonder why it's not $\hbar$ or if there is some additional factor here... In the thermodynamic limit, factors of $2\pi$ become negligible in this expression... $\endgroup$ Apr 22 at 13:39
  • $\begingroup$ Cool thanks very much! That's what I thought, but I just wanted to make sure. $\endgroup$
    – 392or385
    Apr 23 at 10:18

2 Answers 2


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.


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.


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