Fermi energy, occupation factor and number of particles

Question

Using the grand canonical ensemble, we can show that the occupation factor of an energy level (when the temperature $T$ and chemical potential $\mu$ are fixed) is given by $$ f_E(T,\mu) = \frac{1}{\exp \frac{E-\mu}{kT} \pm 1} \quad (1)$$. The total number of particles and energy in the system are thus given by $$ N(T,\mu) = \int dE \, DoS (E) f_E(T,\mu) \quad (2)$$ $$ U(T,\mu) = \int dE \, E\, DoS (E) f_E(T,\mu) \quad (3)$$

On the other hand, the chemical potential is related to the internal, free or Gibbs energy of the system as $$\mu = \left( \frac{\partial U}{\partial N} \right)_{S,V}= \left( \frac{\partial F}{\partial N} \right)_{T,V}= \left( \frac{\partial G}{\partial N} \right)_{T,p} \quad (4)$$

Is there a way to recover these relations from (1) - ie to check that the $\mu$ which appears in eq.(1) is indeed a chemical potential is the sense of (4) ?

Answer 1

Solved it. The grand potential is $$ A =kT\int dE\,D(E)\log\left(1-\frac{1}{\exp\left(\frac{E-\mu}{kT}\right)+1}\right) =-kT\int dE\,D(E)\log\left(1+\exp\left(\frac{\mu-E}{kT}\right)\right)$$

from where we calculate $$ S=-\left.\frac{\partial A}{\partial T}\right|_{\mu,V}=-\frac{A}{T}-\frac{1}{T}\int dE\,D(E)\frac{(\mu-E)}{\exp\left(\frac{E-\mu}{kT}\right)+1}$$

and $$ p=-\left.\frac{\partial A}{\partial V}\right|_{T,V}=-\frac{A}{V}$$

and it is then straightforward to calculate $$G=U-TS+pV =\mu\int dE\,\frac{D(E)}{\exp\left(\frac{E-\mu}{kT}\right)+1} = \mu N$$

CQFD.