Tuesday, July 14, 2009

Thermodynamic (in)stability of hydrogen

The interview season for the "Theoretical and Mathematical Physics" master programme at LMU is approaching quickly. We have to come up with new questions and problems that help us judge our applicants.

It turns out to be easy to find questions in quantum mechanics and those easily lead over to mathematics questions. However, we were always short on good stat mech problems. One possibility is to have an easy start with the harmonic oscillator and then couple that to a heat bath and compute the partition function (with geometric series featuring).

But this time, we thought we could vary this a bit and came to a surprising realization: Hydrogen is unstable! This was news to me but google finds a number of pages where this is discussed. Often wrongly, but the good explanation is in a 2001 paper by Miranda.

The idea is the following: Everybody knows that the energy of the n-th level of the hydrogen atom has energy proportional to 1/n^2. This level has degeneracy n^2 since l runs from 1 to n and m then runs from -l to l. So the partition function is Z(T) = sum_{n=1}^\infty n^2e^{Ry/kTn^2}. First, we thought that this might be a function named after some 19th century mathematician but mathematica told us its name is actually \infty since the exponent quickly approaches 1 for every positive T.

The conclusion seems to be that there is something wrong with the hydrogen atom. And we have not even started to consider the positive energy scattering states. Obviously, this problem has an IR divergence and it is probably better to embed it in some cavity of finite radius. But still, you would think that then for a large cavity, most of the statistical weight would be in the highly excited states and the probability to be in the ground state would go to zero as the cavity gets larger.

The conclusion would be that a hydrogen atom at any temperature would almost never be in its ground state but always highly excited or even ionized. And all this only because the density of states diverges at 0. This looks like a situations worse than the Hagedorn transition that strings experience due to the exponentially growing density of states.

The solution in the above mentioned paper is quite simple: Rather than these scaling arguments one should put in some numbers! Let us start with the Bohr radius, which is 5\cdot 10^{-11}m and the radius grows like n^2. This means in ameter sized cavity we can only fit states up to roughly n=10^5. However, at room temperature, Boltzmann exponent 13.6eV/kT \approx 527 and e^{527}\approx 7\cdot 10^{225}. Thus, to balance the Boltzmann suppression of the higher levels compared to the ground state one has to take into account at the order of 10^{225} states and not just the first 10^5. Or put differently, one should use and exponentially large cavity. Otherwise the partition function is essentially cut off at n=10^5 and the probablility to find the ground state is very very very close to 1.


RZ said...

I don't have access to the paper(is there an arxiv version?) but the result is pretty damn amazing.
I seem to remember an undergraduate calculation showing that earth's atmosphere cannot be in thermodynamic equilibrium. I suppose the calculation is similar. At the time the instructor said that this is because the temperature is not fixed.

Converter said...

The result is amazing .your paper is very informative i like it very much.