# Grand canonical potential for a Fermi-Dirac or Bose-Einstein gas

I have a question about the FD and BE grand canonical potential. We derived the expression for both cases, but probably the expressions must be wrong.

$$J_{FD}=kT\sum_i\ln\left(1+ e^{-\beta(\epsilon_i - \mu)}\right)$$ for Fermi-Dirac.

$$J_{BE}= -kT\sum_i\ln\left(1 - e^{-\beta(\epsilon_i - \mu)}\right)$$ for Bose-Einstein.

But shouldn't it be the other way around, meaning plus sign below and negative sign above?

• Why do you think that? Jun 4, 2021 at 17:41
• because when you derive the grand canonical partition functions, for the bose einstein you have an expression on power -1 which means when you try to find the potential witht he formula -ktlY, you get a 2nd minus, which will result in a plus Jun 4, 2021 at 17:54
• Are you referring to the minus sign before the $kT$, or the minus sign in $1 \pm e^{-\beta(\epsilon-\mu)}$? My initial assumption was that you meant the latter. Jun 4, 2021 at 19:20

The Grand Canonical Partition function look like $$\ln \mathcal{Z}=\pm\sum_i\ln(1\pm e^{\beta(\mu-E_i)})$$ where the $$\pm$$ sign means $$+$$ for fermions and $$-$$ for bosons. $$\Phi_G=-k_BT\ln\mathcal{Z}=\mp k_BT\sum_i \ln(1\pm e^{\beta(\mu-E_i)})$$
• $Y_{FD} = \Pi_i (1+ e^{-\beta (\epsilon_i - \mu)})$ for the fermi dirac grand canonical partition function. $Y_{BE} = \Pi_i (1- e^{-\beta (\epsilon_i - \mu)})^{-1}$ Bose einstein. If you use this expressions, you will get the minus for FD and plus for BE Jun 4, 2021 at 19:08