In statistical mechanics the entropy of a macroscopic system in equilibrium, assuming equipartition, is equal to the logarithm of the number of microstate compatible with it, up to a fixed and well-defined constant. This coincides with the traditional definition of the entropy (which is where the constant comes from).

This definition should be independent of the underlying theory, but how is that possible? When substructure is considered or discovered (e.g. of particles in a gas), or when theories are replaced or refined, the number of microstates (according to the theory) will generally change. It doesn't seem to be the case however that the absolute entropy can not be computed unless a complete theory of every detail of the system is known. Moreover entropy is something that can be measured without having any idea of theory or substructure.

Finally many or most systems have an infinite state space, for example a classical many-particle macrostate will correspond to an infinite subset of the phase space, and an ideal quantum gas at sufficiently high temperature/energy so that unbounded states are accessible will correspond to an infinite dimensional subspace of its Hilbert space. How would this (i.e. the counting of states) work there?

Without a doubt this is a very naive question, but i really have no idea of what the answer could be.

  • $\begingroup$ It would seem that your question boils down to whether or not one should believe that statistical mechanics and thermodynamics can be 'unified', is that correct? Having an infinite state space does not imply that you can access all of it. Theoretically, my wave function extends past the moon, but I don't worry about all of a sudden finding myself there. $\endgroup$
    – Jon Custer
    Apr 7, 2017 at 17:51
  • $\begingroup$ @joncuster I do believe it, I just don't understand how it works. In any case, if that state (you on the moon with nonzero probability) is compatible with the macroscopic state of the system, it should contribute to the count, if I understand correctly. And indeed, having an infinite state space doesn't mean that all states are accessible, but neither is our the case that always only a finite number of states (or in the case of the Von Neumann entropy a finite dimensional subspace) is accessible. $\endgroup$
    – doetoe
    Apr 7, 2017 at 18:01
  • $\begingroup$ The rough answer is that the contributions from high energy states only become relevant at high temperatures. Try to compute the partition function of two quantum oscillators with very different frequency, for example. $\endgroup$
    – Javier
    Apr 7, 2017 at 18:06
  • $\begingroup$ @javier but depending on the system, at any fixed temperature you could have infinite degeneracy. Eg in a classical ideal gas the temperature is proportional to the average kinetic energy. There are infinitely many configurations with the same average kinetic energy, both in classical mechanics and in classical quantum mechanics $\endgroup$
    – doetoe
    Apr 7, 2017 at 18:21
  • $\begingroup$ Temperature is related to average energy per particle, not per configuration. And since entropy is logarithmic, any degeneracy ends up as an irrelevant additive term. $\endgroup$
    – Javier
    Apr 7, 2017 at 19:16

1 Answer 1


If we describe a system using quantum mechanics and then consider the classical regime where we can give an approximate description using a classical phase phase, then what you find is that the density of states per unit phase space volume per particle $d^3xd^3p$ is $\frac{1}{h^3}$. Planck's constant enters in here, which you can interpret as a scaling constant. So, you can indeed choose to "zoom out" so that the phase space description become better and better and everything looks more classical, but the density of states per unit phase spaced volume in your more coarse grained phase space then will increase.

While we can do "classical statistical physics" where we start with a classical phase phase, this is physically not the right way to do statistical physics. Given some well defined physical system you could only have rescaled the coordinates and momenta by some finite amount, so that in terms of the new coordinates the density of states per particle would become $\frac{A}{h^3}$ where $A$ is the Jacobian of the scaling transform. The idea of being allowed to start at the infinite scaling limit and then not having to pay a price for that (like not being able to define the absolute entropy) is flawed.

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    $\begingroup$ Thanks. But Boltzmann's interpretation of thermodynamic quantities predates quantum mechanics, so I find it surprising if Planck's constant plays such a crucial role in counting the number of states. $\endgroup$
    – doetoe
    Apr 24, 2017 at 19:55
  • $\begingroup$ @doetoe Yes, but note the tension between classical physics and thermodynamics as e.g. in the UV catastrophe. The fact that in classical physics Planck's constant is zero will get you in trouble whenever the absolute entropy is needed. Processes where more particles appear, not just in black body radiation, but also evaporation, chemical reactions where on the right hand side if the equation there are more molecules than on the left hand side, you'll find infinite rates. A cup of tea would have to evaporate infinitely fast according to classical physics. $\endgroup$ Apr 24, 2017 at 20:01

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