I came across the following question in some example problems for my graduate statistical mechanics course final exam:
Consider an ideal gas of fermions of density $n$ in three dimensions with the single-particle eigenstate energies given by $\epsilon_k^{\pm}=\pm \hbar c |\mathbf k|$. Assume the chemical potential $\mu = 0$ at $T=0$. Prove that if the fermion density is constant, $\mu(T)=0$ for all $T$.
My first approach was to use the Fermi-Dirac distribution (using energy units for the temperature):
$$N=\sum_k\langle n_k\rangle = \sum_k \frac{1}{e^{\frac {\epsilon_k-\mu}T}+1}=\sum_k \frac{1}{e^{\frac {\hbar c |\mathbf k|-\mu}T}+1}$$
Converting this into an integral in momentum space:
$$N=V\int_{\Bbb R^3}\frac {d^3k}{(2 \pi)^3}\frac{1}{e^{\frac {\hbar c |\mathbf k|-\mu}T}+1}=\frac {V}{2 \pi^2}\int_0^\infty dk \ \frac{k^2}{z^{-1}e^{\frac {\hbar c k}T}+1}$$
Where $z=e^{\frac {\mu}{T}}$ is the fugacity. Using change of variables, we get:
$$N=\frac {V}{2 \pi^2} \left(\frac{T}{\hbar c}\right)^3\int_0^\infty dx \ \frac{x^2}{z^{-1}e^{x}+1}$$
Which in terms of the Fermi-Dirac functions, is :
$$n=\frac {1}{ \pi^2} \left(\frac{T}{\hbar c}\right)^3f_3(z)$$
Where $n$ is the density. At low temperatures, using the Sommerfeld expansion $f_3\left(e^{\frac{\mu}{T}}\right) \simeq \frac 16\left(\frac {\mu}{T}\right)^3\left(1+\pi^2 \left(\frac T{\mu}\right)^2\right)$, we get:
$$n\simeq\frac {1}{ \pi^2} \left(\frac{T}{\hbar c}\right)^3\frac 16\left(\frac {\mu}{T}\right)^3\left(1+\pi^2 \left(\frac T{\mu}\right)^2\right)\simeq \frac {V}{ 6\pi^2} \left(\frac{\mu}{\hbar c}\right)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T\rightarrow0^+ $$
Which gives :
$$\mu \simeq \hbar c\left(6 \pi^2 n\right)^{\frac{1}{3}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T\rightarrow0^+ $$
I can't see how we can assume the chemical potential to be zero at zero temperature. Clearly, the above solution shows that that is impossible. How could the gas have zero fermi energy? I get the same result when calculating the grand canonical partition function from scratch and using the grand potential. A zero chemical potential doesn't make intuitive sense either, because it implies that adding or subtracting particles from the gas would have no energy cost whatsoever; which can't be true, because when we add a fermion it can't share an energy level with another particle (Pauli principle), and thus has to go to a higher energy level, which means that there is a nonzero energy cost.
Another problem is that I can't understand what the question means by the $\pm$ in the $\epsilon_k^{\pm}=\pm \hbar c |\mathbf k|$ relation. Does it mean that each eigenstate of momentum corresponds to two energy eigenstates, one negative and one positive? For example, consider the - case $\epsilon_k=- \hbar c |\mathbf k|$. The same reasoning as above gives:
$$n=\frac {1}{2 \pi^2} \left(\frac{T}{\hbar c}\right)^3\int_0^\infty dx \ \frac{x^2}{z^{-1}e^{-x}+1} \ \rightarrow +\infty$$
And this integral clearly diverges for all values of $z$.
I feel like I'm misinterpreting the question. Any help would be appreciated.
