Why the entropy of non-zero nuclear spin is zero at $T = 0$?

Question

When reading Concepts in Thermal Physics (second edition) by Stephen and Katherine about the concepts of the third law, I met with such a problem. The text reads as follows:

Consider a perfect crystal composed of $N$ spinless atoms. We are told by the third law that its entropy is zero. However, let us further suppose that each atom has at its centre a nucleus with angular momentum quantum number $I$. If no magnetic field is applied to this system, then we appear to have a contradiction. The degeneracy of the nuclear spin is $2I+1$ and if $I> 0$, this will not be equal to one. How can we reconcile this with zero entropy since the non-zero nuclear spin implies that the entropy $S$ of this system should be $S= Nk\ln(2I + 1)$,to however low a temperature we cool it?

Although this question is answered from other aspect at next paragraph:

The answer to this apparent contradiction is as follows: in a real system in internal equilibrium, the individual components of the system must be able to exchange energy with each other, i.e., to interact with each other. Nuclear spins actually feel a tiny, but non-zero, magnetic field due to the dipolar fields produced each other, and this lifts the degeneracy. Another way of looking at this is to say that the interactions give rise to collective excitations of the nuclear spins. These collective excitations are nuclear spin waves, and the lowest-energy nuclear spin wave, corresponding to the longest-wavelength mode, will be non-degenerate. At sufficiently low temperatures (and this will be extremely low!) only that longest- wavelength mode will be thermally occupied and the entropy of the nuclear spin system will be zero.

What I am asking is about the expression $S= Nk\ln(2I + 1)$. Entropy $S$ should be larger than zero for $I>0$. Why we can neglect this expression and say $S=0$ at $T=0$? Is it because $I$ becomes zero or this expression is no more valid at $T=0$?

Answer 1

The expression $S=Nk_B \ln (2I+1)$ works for spins which have no interactions with each other and are not subject to any external field. This is only a good approximation at high temperature. The more general expression is $S= k_B\ln \Omega$ where $\Omega$ is the number of microstates with a given energy. At high temperature, we can ignore all these interactions and any spin state is equally likely, so $\Omega=(2I+1)^N$ and $S=Nk_B \ln (2I+1)$. At $T=0$, interactions between atoms can not be ignored anymore. Usually interactions between spins selects a unique ground state, which means $\Omega=1$ and thus $S=0$.

An instructive exercise is to calculate the entropy of spins in the presence of an external field. In that case $S$ decreases with temperature, reaching $S=0$ at $T=0$ when there is a unique ground state of all spins polarized.

Answer 2

What the answer cited in the OP says, is that in practice the degeneracy is lifted - i.e., the $2I+1$ spin states are no more degenerate, and there is the lowest energy eigenstate among them. At zero temperature only this state is occupied, and hence the entropy is zero.

They make an implicit reference to the manner in which statistical physics deals with interactions - assuming that they are present and thus assure the relaxation to thermal equilibrium, but neglecting them in actual calculations, since they are weaks (hence the use of non-interacting Hamiltonians, like that of an ideal gas without collisions). If the interactions were truly absent, we would not be able to use any of the statistical physics apparatus - in particular, the claim about the entropy being zero at zero temperature would be meaningless.

Such failure of statistical phsyics does happen in real life, when we talk about spontaneous symmetry breaking and phase transitions: the system of interest (e.g., a ferromagnet) finds itself in one specific ground state out of many, because the transitions between these ground states are not possible (at least not on realistic time scales).