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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 $T\to\infty$?

Kagaratsch
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    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 02:24
  • Could you elaborate a little on why your intuition is right? – Kagaratsch Jan 10 '14 at 02:29
  • At any finite temperature, a molecule 'prefers' the lower energy state. – Danu Jan 10 '14 at 02:33
  • 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 02:38
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    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 02:42
  • I think you are misunderstanding what it means to be a Statistical ensemble. The distribution tells you what happens if you take a BUNCH of particles and jitter them about. Some of the particles will be in the ground state. Some will be in the excited state. The distribution doesn't say anything about relaxation rates. that would involve non-equilibrium thermodynamics. – nervxxx Jan 10 '14 at 02:46
  • If the states really are populated equally, this will mean that in a unit of time equal number of particles relaxes to ground state, as well as gets excited into the excited state. That is an equilibrium alright, but it has relaxation constantly going on in the system. And that is what I doubt. – Kagaratsch Jan 10 '14 at 02:50
  • Your statement on the relaxation is correct. And there is nothing wrong with it. Unfortunately, there is nothing more I can say to change your intuition. – nervxxx Jan 10 '14 at 02:55
  • Ok, I just did the computation for a system with three possible states, and indeed the probability for each one at infinite temperature is simply $1/3$. I guess each possible energy state is something like a degree of freedom. At infinite temperature, the system is in a bath of infinite energy and therefore the notion of energy becomes conformal, so that each particle can simply jump through all available degrees of freedom unconstrained. Now I slowly start to believe it, thank you for the discussion! – Kagaratsch Jan 10 '14 at 03:04

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Perhaps the following (which basically involves investigating what happens for a general system with discrete energy spectrum) will help.

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 $n\in\{1,\dots, N\}$.

Now, to complete your intuition, you simply need to understand "why" the partition function looks the way it does, but that's another story all together for which, admittedly, my own intuition isn't the greatest.

joshphysics
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  • @Kagaratsch Sure thing. You may also find the following useful http://physics.stackexchange.com/questions/65165/why-is-the-temperature-zero-in-the-ground-state/65167#65167, http://physics.stackexchange.com/questions/67755/canonical-partition-function-what-assumption-is-at-work-here/67773#67773 – joshphysics Jan 10 '14 at 04:41