# Question on Boltzman statistics and numbering of states

1. Jan 23, 2014

### Sonderval

Consider a monatomic gas of hydrogen (just to make the example as simple as possible) at a temperature T. If I use Boltzmann statistics, I would say that the probability of finding any arbitrary atom at energy E should be proportional to

$g_i e^{-E_i/(k_BT)} / Z(T)$
where $g_i$ is the state degeneracy (number of states at each level).

Thinking about this, I utterly confused myself in two ways, so I hope someone can help me out.

1. Since the states become more and more dense at higher quantum numbers n (only looking at bound states), it seems as if even at small temperatures, most atoms should be in an excited state with very large n: If I use 1Ry as the energy difference between ground state and a highly excited state (which should be a reasonable approximation), I can always find enough states close to the ionisation energy that the summed probability of each of these states is larger than that of the ground state. Obviously I'm making a mistake here, but I have no clue what it is.

2. Consider just the first to shells. Is the ratio of the probabilty of finding an atom in the ground or in the first excited state given by $e^{-(E_2-E_1)/(k_BT)}$ or do I need an additional factor of 4 because the second shell has one s and three p orbitals?

2. Jan 23, 2014

### Staff: Mentor

I don't understand how you get a high population in high-energy states. Even though the degeneracy increases as $n^2$, the exponential factor dominates even at relatively high temperatures, such as the sun's atmosphere (5800 K). Even at that temperature, the ratio of $n=2$ atoms over $n=1$ is of the order of $10^{-9}$.

Yes, you need the factor of 4.

3. Jan 23, 2014

### Sonderval

@Dr Claude
Thanks for the answer. At least #2 assures me that I'm not totally off in how to count states.

Considering question 1, since we have an infinite number of states with energy less than 1 Rydberg above the ground state, I don't see how the exponential factor can dominate that. For simplicity, let's assume that all excited states are at 13,6eV above the ground state, so they all have the same Boltzmann factor.
If the Boltzmann factor has a value of x, I need 1/x excited states to get equal probability for the excited and the ground state.

4. Jan 23, 2014

### Staff: Mentor

You're right, I was thinking of only the first few excited states. If you take into account all states, the partition function is actually infinite.

The "solution" to this problem is that the sum over all states of hydrogen doesn't make sense. It is strictly valid only for non-iteracting particles, and works only if a hydrogen atom is always a hydrogen atom. Highly-excited atoms can be seen as having an electron with a very low binding energy, and a relatively low-energy collision will be sufficient to ionize it. Therefore, highly-excited states are not significantly populated at any temperature.

5. Jan 23, 2014

### Staff: Mentor

... and to handle the unbound states: to have many of them (or even a continuum), you need a single atom in a very large ("infinite") volume, where ionization is indeed the most probable result in equilibrium. With a finite volume, the number of states is lower, and you get the more realistic case of rare ionizations.

6. Jan 24, 2014

### Sonderval

Thanks for the help, but I'm not sure I understand this solution. (Probably I'm overlooking something obvious) If the excited (but still bound) atoms can also ionize completely, how does this skew the occupation numbers in favor of low-energy states? (I understand that the number of states in a finite volume is bounded, but this still increases the - already infinite - number of possible states.) I still don't see how I can apply the Boltzmann formula and get a correct result, with or without the continuum states, if I have to sum over infinitely many states that are close to the ionisation energy.

7. Jan 25, 2014

### Staff: Mentor

The excited states are not proper eigenstates of the energy - they don't exist long enough for the single atom, and they don't exist at all if you look at the whole system.
And their number is finite if the volume is finite.

8. Jan 26, 2014

### Sonderval

@mfb
I'm not sure I accept the first part of your statement - in principle, these states as calculated from the SGL are eigenstates and would be stable if not for external influences. (And the argument would hold for any Eigenstate of the SGL; making every thermodynamic analysis of a QM system - like the Sommerfeld model - very problematic.)

"And their number is finite if the volume is finite."
I think this hits the nail on the head: In a finite volume, there is no "n=1googolplex"-state because this would be extremely extended. And thinking further, I would assume that for the same reason we cannot really excite these states because the electron would have to delocalise extremely. So even in an infinite volume, to excite a single hydrogen atom into one of these states would probably be very difficult - a perfectly tuned photon could do the trick, but the energy density of all photons in the required energy range (13,6eV above ground state) would be finite.
And thinking further on the problem, the same holds for the higher angular-momentum states of these levels - there is simply no mechanism to excite a hydrogen atom in its ground state into the n=1000000 level with angular momentum number l=999999.

9. Jan 26, 2014

### Jano L.

There is infinity of eigenfunctions near ionization (energy = 0) if the hydrogen atom is the only thing in infinite space.

If we enclose the hydrogen atom in a box of finite volume, only first few eigenfunctions will resemble the the same number of previous ones; gradually, as $n$ increases, the eigenfunctions will begin to resemble those of a free particle in a box. The number of the latter is still infinite, but this time, they are well spaced on the energy axis as $n^2$ with increasing gaps, which makes the partition function $Z$ to converge. There is no longer infinite density of eigenfunctions anywhere on the energy axis - the box removes this.

Now, the box does seem artificial, but it can be estimated that the resulting probabilities for first few eigenfunctions depend very weakly on the volume of the box, so it does not matter whether you take $V^{1/3}$ equal to the mean inter-atomic distance or its twice.

10. Jan 27, 2014

### Sonderval

@Jano
Nicely explained, thanks.