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Suppose I have a perfect crystal with statistical entropy $S_1$. If I cool it to 0 Kelvin, the number of possible microstates becomes 1, hence the statistical entropy becomes 0.

However, statistical entropy is only defined up to an additive constant. This is because there are actually infinitely many microstates, and you need to choose $\delta x$ and $\delta p$ in the course graining process.

Therefore, we could define the initial entropy in another way to see that it is a different number $S_2$. Now with this alternative choice of microstates, the entropy does still tends to zero when we let the temperature go to zero Kelvin.

How much entropy has left the system in the cooling process? Differences in entropy should be independent from the choice of microstates. However, it seems that $\Delta S$ is both $-S_1$ and $-S_2$, which is impossible?

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  • $\begingroup$ The first paragraph contradicts the second one - you seem to be using two different definitions of entropy (perhaps one from classical and the other from quantum statistical physics?) $\endgroup$
    – Roger V.
    Commented Oct 20, 2022 at 7:55
  • $\begingroup$ In the second paragraph, I used the information from the first paragraph of the section ‘Counting of microstates’ of the following Wikipedia page: en.m.wikipedia.org/wiki/Entropy_(statistical_thermodynamics) $\endgroup$
    – Riemann
    Commented Oct 20, 2022 at 8:12
  • $\begingroup$ For the first paragraph of my question: this follows from the Gibbs entropy formula. $\endgroup$
    – Riemann
    Commented Oct 20, 2022 at 8:13
  • $\begingroup$ I think that both are classical $\endgroup$
    – Riemann
    Commented Oct 20, 2022 at 8:13
  • $\begingroup$ Yes, but here you count them in classical phase space (coordinates and momenta) - and in classical statistical physics the entropy is indeed defined up to a constant - which is precisely due to the arbitrariness of choosing a phase cell / physically small volume. On the other hand, in quantum statistical physics the states are labeled by quantum numbers and at zero temperature we truly have only one ground state, and only one term in the statistical sum survives. $\endgroup$
    – Roger V.
    Commented Oct 20, 2022 at 8:15

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I think you are confusing different notions of entropy here. The statement that the entropy is defined only up to a constant is true for the thermodynamic (or Clausius) entropy defined as $$d S_{C}=\frac{\delta Q}{T}$$ For the statistical (or Boltzmann) entropy defined as

$$S_{B}=k_B\log W$$

the number of microstates might be hard to count explicitly at times, but fundamentally it is an objective number, hence there is no ambiguity in $S_B$. Your consideration about arbitrariness in choosing the size of the unit cell in phase space, is in fact from the pre-quantum era when people just didn't know how to count states consistently. With the advent of Quantum Theory, this ambiguity was lifted as well.

As for the entropy at $T=0$, you should know that any system finds itself in its ground state at zero temperature. For a generic quantum system the ground state is normally unique, hence Nernst's theorem about $T=0$ entropy. This fact that the ground state is unique has to do with that fact that a system always tries to get rid of degeneracies. On the other hand in some situations this degeneracy might be protected by some considerations, such as symmetry. In this case the ground state remains degenerate, and you arrive at what X.-G. Wen calls topological order.

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