# On the statistical occupancy of energy levels for a Boltzmann distribution at high temperatures

Suppose we have a two-level thermodynamical system:

In what follows, we will adopt the following convention: $$E_a=0$$, $$E_b=\cal E$$. For a system of $$N$$ classical (non-interacting) particles, the mean total energy is $$U=N\langle E\,\rangle$$, where $$\langle E\rangle$$ is the expected value of the energy of each particle, which is given by $$\langle E\rangle = \sum_{i}E_i\,p_i$$ as prescribed by statistics. In our case $$i=a,b$$ while $$p_i=p(E_i)$$ follows the Maxwell-Boltzmann distribution: $$p(E_i)=\frac{\exp\left(-\frac{E_i}{k_BT}\right)}{Z}, \qquad Z=\sum_i \exp\left(-\frac{E_i}{k_BT}\right)$$

It's easy to see that $$\lim_{T\to\infty}U(T)=\frac{N\cal{E}}{2}$$ meaning at high temperature half of the particles are expected to be found on the "excited" state $$E_b$$, and half of them of the "ground" state $$E_a$$.

Mathematically, this is the right result, but... why does this have to be like this? Intuitively, I expected all the particles on the excited state.

My professor said "no, this can only be the case with negative temperatures", making this whole concept even more obscure.

Particles in the environment have, on average, energy of about $$k_{\textrm{B}}T$$. If $$k_{\textrm{B}}T \gg \mathcal{E}$$, the two internal states of the system are essentially degenerate, as far as the environment is concerned. In other words, it costs essentially no energy (relative to $$k_{\textrm{B}}T$$) for the system to change state. Therefore, in each "collision" of a particle in the environment with a particle in the system, the outcome of the collision is essentially random: the system particle will be in either of its internal states with equal probability.