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

Question

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.

Answer 1

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.