# Why $\epsilon > \mu$ for Bose-Einstein distribution (but not for Fermi-Dirac)?

For fermions $$\bar{n}_{FD}=\frac{1}{e^{(\epsilon -\mu)/kT}+1}$$ and $\epsilon$ can be bigger or small than $\mu$. However, for bosons: $$\bar{n}_{BE}=\frac{1}{e^{(\epsilon -\mu)/kT}-1}$$ which implies that $\epsilon >\mu$ otherwise the occupancy is negative.

Is there another argument as to why $\epsilon >\mu$ must be true for bosons but not for fermions? I am reading Schroeder and he says

It would be negative if $\epsilon$ could be less than $\mu$, but we've already seen that this cannot happen.

but I cannot find where "we've already seen that this cannot happen"

The chemical potential is sort of the potential energy needed to add another particle from the surrounding reservoir to the system. Thus to add another particle to a particular single particle level requires energy if the chemical potential is larger than the energy of single-particle level.

If the chemical potential was smaller than the energy level, then one would lower the systems total energy by adding another particle from the reservoir to that energy level, so one would expect that such a system just collapses. This is also consistent with the fact that the occupation number goes to infinity when the $\mu$ approaches $\epsilon$.

For fermions, this collaps can not happen, since we are constrained by the pauli principle so we can not have an arbitrary number of particles in the same state.

• I found what I was looking for but thanks for the answer. Yours helped with the intuition behind it. – user1830663 Apr 23 '15 at 2:24

I found it:

$$\bar{n}_{BE}=-\frac{1}{\mathcal{Z}}\frac{\partial \mathcal{Z}}{\partial x}$$ where $x =(\epsilon -\mu)/kT$

$$\mathcal{Z} = \sum\limits_{n}^\infty e^{-n(\epsilon -\mu)/kT}$$ which converges if $\epsilon -\mu<0$