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Does the entropy of ideal fermi gas go to zero , in accordance with third law of thermodynamics?

Consider a system of three fermions in a 3D box. The first fermion goes to the ground state of the box. Second fermion and the third will go to the first excited state as the first excited state of the box in triply degenerate.

This will be the ground state of the system, but the entropy of this ground state is not zero because two fermions can be distributed among three states in three ways.From the Gibbs definition of entropy, we have three micro states for the ground state of this system and hence entropy $=k_B\ln3$.

If we had $N$ fermions and we are adding a new fermion to the system this fermion can occupy any state on the surface of the fermi sphere. In this case also the ground state is degenerate and hence the entropy is non-zero.

The chemical potential is defined as the change in internal energy w.r.t change in particle number while keeping the volume and entropy constant.

In the fermi gas case we cannot add a particle to the system while keeping entropy constant. So how can we conclude that the fermi energy is the chemical potential at $T=0$?

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  • $\begingroup$ Usually the third law is given in a form that the entropy per particle should vanish, more precisely there should be no extensive contribution $S*(N)=N\lim_{N'\rightarrow \infty} S(N'}/N'=0$ $\endgroup$ – Bort Feb 22 '16 at 9:50
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Fermions are indistinguishable particles, hence you cannot distibute the three fermions in different ways. The state is completely described by the occupations numbers $|1,2\rangle$ and has zero entropy. If you are talking about a Fermi system at a given temperature, that means it is (or was) in contact with the external world at that temperature (canonical or grand canonical ensemble). The state of such system is described by a density matrix. In equilibrium it is given by $$\rho = \frac{e^{-\beta H}}{Tr \, e^{-\beta H}}.$$ In the energy eigenbasis, the numerator reads $$\sum_n e^{-\beta E_n} \, |E_n\rangle\langle E_n|. $$ In the limit $T\rightarrow 0$, $\rho$ approaches $|E_0\rangle \langle E_0| $. This density matrix describes a quantum mechanical pure state, which always has zero entropy.

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@curio hit the nail on the head regarding indistinguishability of identical particles. As for degeneracy, if there is a degenerate many-particle ground state then as you noted correctly there is nonvanishing entropy at zero temperature. Finally, your remark on not being able to hold entropy constant is addressed in https://en.m.wikipedia.org/wiki/Chemical_potential. There they explain how different expressions for the chemical potential are useful for different problems.

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