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?

  • 2
    $\begingroup$ Check out this paper for some intuitive understanding of the chemical potential for Bose gases (and Fermi gases, and ideal gases). It's usually behind a paywall, but at the moment, you can get free access to these journals if you create an account. $\endgroup$
    – march
    Commented May 20, 2020 at 20:39
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    $\begingroup$ As explained in the paper that @march cited, the relationship $dE=T\,dS-p\,dV+\mu\,dN$ says that $\mu$ describes the change in $E$ when $S$ and $V$ are fixed. Are you taking care to keep the entropy fixed when you calculate the change in energy due to adding a particle (like you mentioned near the end)? In particular, are you using this condition to constrain the $\Delta n_i$? $\endgroup$ Commented May 20, 2020 at 21:39
  • $\begingroup$ In short: if $\mu$ is negative it means that the internal energy is such that $E(N,V,S)< E(N+1,V,S)$. Why? Because $N \rightarrow N+1$ increases the microstates, hence $S$, so to keep $S$ constant while adding a particle you should decrease $E$. Not a general rule, it depends on how much $S$ increases, and this depends on the statistics (for Fermi it can be both positive or negative). See also: physics.stackexchange.com/q/7470/226902 physics.stackexchange.com/q/7621/226902 $\endgroup$
    – Quillo
    Commented Mar 23, 2023 at 10:05

1 Answer 1


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 $\mu \leqslant 0$.

  • $\begingroup$ So your statement is that we cannot change $N$ with retaining the same $S$ in the $(N,V,S)$ ensemble? It seems very strange to me because in this case $N$ and $S$ cannot be treated as independent thermodynamic variables, so the derivative $(\partial E/\partial N)_{S,V}$ is undefined and the $(N,V,S)$ ensemble does not exist at all. $\endgroup$ Commented May 21, 2020 at 18:20
  • $\begingroup$ You probably can change $N$ while keeping $S$ fixed, but then you need something else on top of adding particles, to compensate for this. Hence you cannot just easily identify the chemical potential as the change of internal energy per added particle. $\endgroup$ Commented May 21, 2020 at 18:33

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