I was trying to compute some stat mech and thermodynamic quantities using the data in the HITRAN molecular data base and ran into a conceptual problem. The basic quantity needed for these calculations is the partition function, $Q(T)$, the usual sum over Boltzmann factors. HITRAN provides the spectral line wavenumber, $\nu_{ij}$, and the lower state energy of the transition, $E_{\text{lower},\,ij}$ for every spectral line. I naively computed

$$Q(T) = \sum_{I,j} \exp( - (\nu_{ij} + E_{\text{lower},\,ij})/kT )$$

which gives the wrong answer compared to the the HITRAN tables of precomputed Q(T) values. It turns out that the correct calculation is

$$Q(T) = \sum_{\text{unique }E_{\text{lower},\,ij}}\exp( - E_{\text{lower},\,ij}/kT )$$

With this definition I got the right answer for $Q(T)$ and the computed specific heats of H2O and CO2 matched their measured values within 1%.

What I learned from this is that there are many more spectral lines than there are values of $E_\text{lower}$ and each value of $E_\text{lower}$ has a specific set of lines associated with it. This is true for N2, O2, CO2, and H2O.

A hydrogen atom only has one $E_\text{lower} = 0$ and all its energy levels are with respect to this. Where do the $E_\text{lower}$ values come from in a molecule? Why does the partition function only depend on them and not all the spectral lines?

  • $\begingroup$ A hydrogen atom can transition from $n=3$ to $n=2$. In that case, would level $2$ count as one possible value of $E_{lower}$? $\endgroup$ – Display Name Jan 2 at 21:02

Because every energy state can be excited into a higher one, every energy state that the molecule has will appear as one possible $E_{lower}$. All the partition function cares about is the list of possible energy states, and a convenient list that happens to have all of the same things on it is the list of possible lower states in transitions. The properties of the spectral line itself are typically ignored, because in the thermal limit even a very rare transition will happen often enough to equilibrate that specific energy level.

  • $\begingroup$ None of the E_lower levels appears as the energy of a spectral line. Is it physical or just a book keeping technique for HITRAN? $\endgroup$ – Paul in Boston Jan 3 at 16:25
  • $\begingroup$ A further question. All energy levels in hydrogen are derived from E_lower = 0 with n=0. What's the partition function then? $\endgroup$ – Paul in Boston Jan 3 at 17:11
  • $\begingroup$ I think that there should also be an E_lower for the transition down to n=2. After all, that's a valid transition. I think what we are looking at here is a book keeping idiosyncrasy of HITRAN, because in the normal use of the partition function all you need is a list of the energy levels: the transitions do not matter at all. The energy of a spectral line probably corresponds to the difference in energy between the two states. However the thermodynamics doesn't worry about the differences in energy, needing only the list of absolute energies. $\endgroup$ – Display Name Jan 3 at 17:14

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.