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In my course, there is this fact :

In a Bose gas, the chemical potential $\mu$ must always be lower than the smaller level of energy $\epsilon_0$.

I find this strange, because if we put a Bose gas in a big container of $\mu > \epsilon_0$, what happens ?

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There simply doesn't exist any container with $\mu\gt \epsilon_0$; that's what the quoted sentence says. What you could try is to try to increase the chemical potential. But the Bose-Einstein distribution says $$ \langle n_i\rangle \sim\frac{g_i}{e^{(\epsilon_i-\mu)/kT}-1} $$ and if you chose values $\mu\gt \epsilon_i$, then the exponent in the denominator would be negative which means that the exponential would be smaller than one and the denominator (the exponential minus one) would be negative, therefore implying that the number of particles in the $i$-th state has to be negative. But there are no states with a negative number of bosons in a state so this is just impossible.

If you try to raise $\mu$ towards some $\epsilon_i$, the exponential in the formula for $\langle n_i\rangle$ will converge towards one which means that the denominator will go to zero and $\langle n_i\rangle$ will go to infinity. You won't be able to "surpass" the $\mu=\epsilon_i$ level much like you can't surpass the speed of light. As you approach $\mu\to\epsilon$ from below, it becomes harder and harder to increase the chemical potential further.

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  • $\begingroup$ Are you certain you don't mean $\mu > \epsilon_0$ in the first line? $\endgroup$ – joshphysics Jun 6 '13 at 19:26
  • $\begingroup$ I don't understand : if we create a container with $\mu > \epsilon_0$, and then we put the Bose gas into the container, what happens ? $\endgroup$ – Arnaud Jun 9 '13 at 7:55
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This quote is only for free Bose gas.

For interacting Bose gas, of course the chemical potential can be above the single particle ground state. This is common in the cold atom Bose-Einstein condensates.

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You cannot "create" a container with $\mu > \epsilon_0$. $\mu$ depends on the number of particles in the system as well as the volume. So to explain things a different way than what others said, if you create a container, and keep adding particles, it will increase $\mu$ of the system. However, you can add infinitely many particles before the system reach the condition $\mu = \epsilon$. This is due to the fact that, (as others explained,)

$n_0 = \frac{1}{e^{(\epsilon_0-\mu)/\tau}-1}$

And when you keep adding particles, they occupy energy levels depending on the Bose-Einstein distribution. However, there is a maximum number of particles that can occupy the exited energy states and still obey BE statistics. All the particles in excess gets added to the ground state creating a BEC. Hence you can add an infinite number of particles without increasing the $\mu$ beyond $\epsilon_0$

If you look at the above equation, you will see the number of particles in the ground state with energy $\epsilon_0$, $ n_0 \rightarrow \infty$ as $\mu \rightarrow \epsilon_0$.

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If you do that, particles would move from the system with smaller chemical potential to the system with greater one and two systems have same chemical potential(lower than single particle ground energy) finally. You should note that,a system that have very few particles will have a very lower chemical potential and be smaller than any single particle ground states of any systems of ideal gas. Surely, the chemical potential of ANY system would be always smaller than (single particle) ground energy of container A, if systems REACH thermodynamic EQUELIBRIUM. For reaching thermodynamic equelibrium. First, systems could EXCHANGE particles. Oh!!! Zeroth, systems have same kind of bosons... Else, two system will have different chemical potential.

I wish you understand my English(I name it Lianglish).

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