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Chemical potential of noninteracting bosons is known to be negative because the Bose-Einstein distribution $[e^{(\epsilon-\mu)/T}-1]^{-1}$ should be free of singularities. However, I don't fully understand how it is connected with the thermodynamic definition $\mu=\partial E/\partial N$ of chemical potential as the minimal energy required to add an extra particle to the system. Consider the mean occupation numbers $\{n_i\}$, then $$ N=\sum_i n_i,\qquad E=\sum_i \epsilon_in_i\quad\mbox{and}\quad N'=\sum_i n'_i,\qquad E'=\sum_i \epsilon_in'_i $$ before and after adding an extra particle. If $\Delta n_i=n_i'-n_i$, then $$ \Delta N\equiv N'-N=\sum_i\Delta n_i,\qquad\Delta E\equiv E'-E=\sum_i \epsilon_i\Delta n_i. $$ Since $\Delta N=1$, we have $$ \Delta E\geqslant\min(\epsilon_i)\sum_i\Delta n_i=\min(\epsilon_i)\geqslant0. $$ So we obtain $\Delta E\geqslant0$ hence $\mu$ should not be negative.

One might argue that we should work in canonical ensemble where $\mu=\partial F/\partial N$, $F$ is the free energy. However we can consider, for example, the $(N,V,S)$ ensemble, where $dE=TdS-pdV+\mu dN$, so $\mu$ is still $\partial E/\partial N$. I suppose it is possible to add an extra particle in such a way that $S$ will not change.

So why the chemical potential of noninteracting bosons is negative despite the fact that nonnegative energy is needed to add a particle to the system?

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Alexey Sokolik
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1 Answers1

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You are correct that $\mu = \partial U/\partial N$, but the partial derivatives mean that:

$$ \mu = \left (\frac{\partial U}{\partial N} \right )_{S, V},$$ i.e. constant entropy.

For quantum degenerate gases, where state occupancy and number of particles are comparable, you cannot assume that your entropy is staying the same.

A better point to start from is the free energy $F = U - TS$, from which you can define:

$$ \mu = \left (\frac{\partial F}{\partial N} \right )_{T, V}, $$ so that entropy need not be constant.

In this context, essentially, the chemical potential $\mu$ is the change in Helmholtz free energy when a particle is added to the system.

Adding a particle at a particular temperature increases the internal energy $U$, but this extra particle also results in many more possible arrangements of the particles in the system, which in turn increases the entropy $S$.

For bosons nearing the BEC transition: in the thermal phase, the entropy change is larger than the energy term, hence the chemical potential is negative $\mu < 0$.

This agrees with the "mathematical" argument of making sure the bosonic occupancy function stays physical, i.e. $f(E) > 0$. The occupancy of bosons $$f(E) = \frac{1}{\mathrm{e}^\frac{E-\mu}{k_{\mathrm{B}}T}-1}$$ has to be positive, which means that $E-\mu \geqslant 0 \quad \forall E$. So you if fix $\mu$ and choose your ground energy to be $E_0 = 0$, then $\mu \leqslant 0$.

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