$$p_i = \frac{ \exp\left(-\frac{\epsilon _i}{k_BT} \right)}{Z} $$ $$ Z= \sum_{i} \exp\left(-\frac{\epsilon _i}{k_BT} \right)$$

A) Is $p_i$ the probability of the system having an energy equal to $\epsilon_i$? (Probability to be in any of the many microstates that have energy $\epsilon_i$).

B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $\epsilon_i$? (This microstate is not the only microstate with the same energy).

If A) is correct then: $$ Z= \sum_{\epsilon_i} \exp\left(-\frac{\epsilon _i}{k_BT} \right)$$

If B) is correct then: $$ Z= \sum_{\epsilon_i} \Omega_i\exp\left(-\frac{\epsilon _i}{k_BT} \right),$$ where $\Omega_i$ is the multiplicity of the macrostate of energy $\epsilon_i$.

From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.

What is the correct interpretation of the Boltzmann distribution?


1 Answer 1


To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $\varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $\varepsilon_i$.

The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=\sum_i e^{-\frac{\varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.

  • $\begingroup$ I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(\epsilon)$ states. $\endgroup$ Commented Jan 7, 2019 at 21:09
  • $\begingroup$ I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement. $\endgroup$ Commented Jan 7, 2019 at 21:10

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