# Interpretation of the Boltzmann factor and partition function

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

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?

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.
• 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. – By Symmetry Jan 7 at 21:09