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In equilibrium statistical physics the fundamental assumption of statistical thermodynamics states that the occupation of any microstate is equally probable (i.e. $p_i=1/\Omega, S=-k_B\sum p_i\,{\rm ln}\,p_i=k_B{\rm ln}\,\Omega$). But for isolated system in equilibrium we also have Boltzmann distribution which states $p_i=e^{-\beta E_i}/Z$, where $E_i$ are the allowed energy levels. So the two $p_i$ matches if and only if there is one single allowed energy level. How can we resolve this conflict?

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The equal probabilities are meant for states of an isolated system with constant total energy. Each state with this energy is then equally probable.

The Boltzmann probabilities are meant for systems in thermal contact and equilibrium with reservoir of definite temperature - in that case the energy of the system may change due to interaction with the reservoir, so it makes sense that probability of state is correlated with its energy.

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  • $\begingroup$ I edited the wording to try to make the answer more clear. If you don't like it you can just roll back the edit. $\endgroup$ – DanielSank Dec 28 '14 at 7:09
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To clarify, the full assumption is that all states are equally probable, in the absence of any knowledge about the state of the system.

With the Boltzmann distribution (AKA canonical ensemble) this assumption doesn't apply since we have knowledge about the system. In particular we know that if the system is put into contact with a thermodynamic heat bath of temperature $T$ (the same $T$ as in the $e^{-E/(kT)}$ distribution), the system will remain in statistical equilibrium (the distribution will not change). This property, of being in equilibrium with other systems of the same temperature, is special to the Boltzmann distribution and is what makes it so useful.

It's worth noting that many texts give a deceptive statement of the fundamental assumption, that "for isolated systems all states are equally probable". This is not true - isolated systems can be in any distribution. For example if a system is in the Boltzmann distribution because of contact with a heat bath, and that contact is removed to make the system isolated, then its state continues to be described by the Boltzmann distribution. Only after new information about system's state is obtained, does its probability distribution change to something else.

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