Boltzmann distribution and the most probable configuration

Question

According to boltzmann distribution, $$\frac{N_i}{ N_j}=e^{-\beta(\epsilon_i - \epsilon_j)}$$

Where $N_x$ is number of particles in the $x^{th}$ state in the most probable configuration. $\epsilon$ is energy of that state and $\beta$ is temperature dependent constant.

So, from this we can infer that a higher energy state will always have lesser number of particles than a lower one in the most probable configuration, right?

(For convenience, let us assume that every energy level has only one state.)

My issue is, couple small examples I take seem to violate this principle.

For instance, take the case where you have 5 particles with total energy $5\epsilon$in a system with energy levels $0,\epsilon, 2\epsilon...$. You chalk out a table to find weights of all possible configuration.

Image from Atkins solutions

Now, the most probable configurations are {2,2,0,1,0...} and {2,1,2,0,0...}. But both of these cannot be the most probable configuration if Boltzmann's law is true.

Where am I going wrong?

Answer 1

Firstly, the Boltzmann distribution describes the entire ensemble, not specific microstates.

Secondly, the Boltzmann distribution is irrelevant here. The microstates that are listed in your book belong to the microcanonical ensemble, which the Boltzmann distribution does not apply to (it applies to the canonical ensemble).

Answer 2

The Boltzmann distribution only holds for really large N and macroscopic amounts of energy. You can't hope to have it work for just 5 particles and 5 quanta. The Boltzmann distribution assumes approximations that are actually very close to exact, but only if the number of particles of the system is uncountably large.