0

For a system of interaction free identical fermions/bosons in thermodynamic equilibrium, the average number of particles in a single-particle state $i$ is given by, $$n_i=\frac{1}{e^{\beta (\epsilon_i-\mu)}\pm1 },$$ where $k$ is Boltzmann's constant, $T$ is the absolute temperature, $\epsilon_i$ is the energy of the single-particle state $i$, and $\mu$ is the chemical potential.

Is it true, that the average number $\bar{n}_i$ of the corresponding antiparticles is given by a sign-flip of the chemical potential ($\mu\to-\mu$) in the above formula?

  • See https://physics.stackexchange.com/questions/7470/chemical-potential and links therein. – Alpha001 Aug 11 '18 at 16:34
  • Thx. However, from that link it becomes not clear if the energy stayes the same. Also the explaination concerning the relativisitc case seem somewhat difficult to understand or incomplete to me. –  Aug 11 '18 at 16:56

0 Answers0