# 'Fermi-Dirac'-like occupation probability at high temperature

Consider an ensemble of $N\to\infty$ free particles, each of which can assume energy states $E_i\in\{0,E\}$. Using the canonical ensemble one can compute the occupation probability for a single of those particles to be in the excited state $E_i=E$ (or equivalently the expectation value for what fraction of all particles is in the excited state). The result is:

$$n_T(E)=\frac{1}{e^{\frac{1}{k_B T}E}+1}$$

Now, if we check this expression in the limit $T\to 0$, we properly obtain $n_0(E)=0$, telling us that at low temperatures almost no particles will be in the excited energy state. But then, in the opposite limit $T\to\infty$ we get $n_\infty(E)=1/2$, so apparently at infinite temperature there will be equally many particles in the ground and the excited state! I kind of feel like all the particles should go into the excited state for $T\to\infty$, so that this goes against intuition. But maybe I am wrong? What should I expect to happen for $T\to\infty$?

• The probability of occupying the ground state is always greater than the probability of occupying the excited state, regardless of the temperature. In the high temperature limit both become equally likely. In short, your intuition is wrong. – nervxxx Jan 10 '14 at 2:24
• Could you elaborate a little on why your intuition is right? – Kagaratsch Jan 10 '14 at 2:29
• At any finite temperature, a molecule 'prefers' the lower energy state. – Danu Jan 10 '14 at 2:33
• Hmm, maybe there is a thermodynamical rule or something? I still don't see how a particle could possibly relax into the ground state when there is infinite temperature going crazy all around it. – Kagaratsch Jan 10 '14 at 2:38
• The statements I made follow simply from your formula. At the risk of being tautological, let me present some intuition: a system prefers to be in the lowest possible energy level. That is very reasonable, no? Like how water always flows from the highest point to the lowest point. Now temperature just jitters the system around, allows it to POSSIBLY populate higher levels. What that means isn't that the ground state is now inaccessible, it just means that now the system can choose between all possible levels, and it is reasonable that each one of the levels can be occupied with equal prob. – nervxxx Jan 10 '14 at 2:42

The canonical partition function for a quantum system with discrete spectrum $\{E_n\}$ is \begin{align} Z = \sum_ne^{- E_n/(kT)} \end{align} and the population fraction of the systems in the ensemble with energy $E_n$ is given by \begin{align} p_n = \frac{e^{-E_n/(kT)}}{Z}. \end{align} Now, consider two energy levels $E_n$ and $E_m$, then the ratio of their population fractions is \begin{align} \frac{p_n}{p_m} = e^{-(E_n - E_m)/(kT)} \end{align} Now here's the key point. As long as the difference $E_n-E_m$ is finite (which of course it will be for any two energies in the spectrum), the $T\to\infty$ limit of this expression always gives $1$! The difference in the energies gets "washed out" by the largeness of $T$. This tells us that at high temperature, any two levels in the spectrum will have an equal likelihood of being populated!
In particular, if the system has a finite-dimensional Hilbert space, say of dimension $N$, then the probabilities must add to $1$ and must all be equal in the high-temperature limit; \begin{align} p_1+p_2+\cdots+p_N = 1, \qquad \text{$p_n = p_m$ for all $m,n\in\{1,\dots,N\}$} \end{align} which gives $p_n = 1/N$ for all $n\in\{1,\dots, N\}$.