Statistical mechanics and Planck constant universality

Question

In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.

Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.

To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.

In the grand canonical ensemble, the key potential is the grand potential, which is $$\Phi = U - TS-\mu N .$$ The link to statistical mechanics comes from $$\Phi = -kT \ln(\mathcal{Z}), $$

where $$\mathcal{Z} = \sum_N e^{\beta \mu N} Z(V,N,T).$$

For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that

$$Z = \frac{1}{h^{3N} N!} \Big(\int d^3p d^3q \, e^{-\beta E({p}, {q})}\Big)^N$$

This yields in turn that

$$\mathcal{Z} = e^{\frac{1}{h^3} \Big(\int d^3p d^3q \, e^{-\beta E({p}, {q})}\Big)e^{\beta \mu}}$$

and in turn that $\Phi$ is proportional to $\frac{1}{h^3}$:

$$\Phi = -kT \frac{1}{h^3} \Big(\int d^3p d^3q \, e^{- \frac{E({p}, {q})}{kT}}\Big)e^{\frac{\mu}{kT}}. $$

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For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $\mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have

$$\Phi = -\frac{16 \pi k^4}{h^3 c^3} T^4 V.$$ This leads to a pressure

$$P = \frac{16 \pi k^4}{h^3 c^3} T^4$$ from the thermodynamic relation above.

As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $\mu$.

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I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?

Answer 1

I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.

I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.

I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.

Answer 2

The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.

In particular, the chemical potential contains a $\ln(h^3/\text{stuff})$ factor (e.g. the classical ideal non-relativistic gas has $\mu \propto \ln\left(\frac{\lambda^3 N}{V}\right)$, with $\lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.

Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $\mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.