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If the chemical potential of a fermionic system is $0$ at temperature $T=0$, will it be zero at any arbitrary finite temperature?

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  • $\begingroup$ Can you explain why that should be the case? The zero-point of a scale measuring energies like the Fermi energy is pretty arbitrary no? $\endgroup$
    – Marius Ladegård Meyer
    Commented Feb 29 at 17:26
  • $\begingroup$ No, at finite temperature it will move. However, if the chemical potential of a fermionic system is zero, then either it is not a conserved quantity and can be created and destroyed with impunity (then its chemical potential is strictly zero always, however, this is extremely unlikely: we have no example of a fermion that behaves this way), or there are no fermion particles, in which case the chemical potential is not really well-defined. Your situation seems to be moot. $\endgroup$
    – naturallyInconsistent
    Commented Feb 29 at 17:53
  • $\begingroup$ Say the density of states of a fermionic system is given to be [ g(E) = \begin{cases} 0 & \text{if } E < -E_0 \\ \frac{2n_0}{E^2_0} |E| & \text{if } -E_0 \leq E \leq E_0 \\ 0 & \text{if } E > E_0 \end{cases} ] where $E_0 (>0)$ is a given energy, and $n_0$ is number density of the fermions. This gives us the chemical potential at $T=0$ to be $0$. But I wonder what happens at any arbitrary finite temperature. $\endgroup$
    – Snpr_Physics
    Commented Feb 29 at 21:00

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