In an old paper, Ehrenfest 1931, the introduction starts off as follows:

The band spectra of symmetric diatomic molecules show certain striking differences from those of asymmetric molecules. For when the two nuclei of the molecule are identical, the intensity of the individual lines of a band, instead of varying smoothly from line to line, alternates more or less markedly.

There is something called the relative weight, defined by the ratio of the number of states that are symmetric under interchange of the two nuclei and the number that are asymmetric. In modern language, let the degeneracy of the nuclear ground state be $g$. Then the relative weight is $(g+1)/(g-1)$ if the nuclei are bosons, $(g-1)/(g+1)$ if fermions. If I'm understanding correctly, then if the rotational state of the molecule is $J$, then depending on whether $J$ is odd or even, the two nuclei can pick up a relative phase $(-1)^J$, so the degeneracy of the odd-$J$ and even-$J$ states alternates according to the relative weights.

Assuming the above is right, then there's only one thing I really don't understand. How does this show up experimentally as an alternating pattern of intensities? What kind of transitions are we talking about? M1? E2? Are these absorption spectra? Emission spectra? Suppose you populate a molecular state with some spin. Then we're just hopping down a ladder, and it seems to me that by simple conservation of probability there can't be any alternation of intensity. Or by intensity do they actually mean transition rate rather than what an experimentalist would call intensity? (If so, then I'm curious how one would determine these transition rates. From natural line widths? From competition between radiation and collisions in a sample of gas?)

P. Ehrenfest and J. R. Oppenheimer, "Note on the Statistics of Nuclei," Phys. Rev. 37 (1931) 333, http://link.aps.org/doi/10.1103/PhysRev.37.333 , DOI: 10.1103/PhysRev.37.333

  • $\begingroup$ One point to clarify: since symmetric molecules have zero dipole moment, they do not interact with electromagnetic radiation. Therefore, purely rotational transitions are forbidden. What you can observe are transitions to another electronic states, together with a ±1 change in J. $\endgroup$ – gigacyan Sep 10 '13 at 20:39
  • $\begingroup$ @gigacyan: Hm...but can't you, e.g., get electric quadrupole radiation between rotational states with the same electronic state? $\endgroup$ – user4552 Sep 10 '13 at 21:15
  • $\begingroup$ yes, but selection rules will be different: quadruple-allowed are J±2 transitions (in Raman spectroscopy, for instance). $\endgroup$ – gigacyan Oct 15 '13 at 12:58

This is a real separation of molecules according to mutual orientation of both nuclear spins. For example, the energy difference of ortho- and parahydrogen is sufficient to boil 50% of liquid hydrogen as orthohydrogen changes slowly to parahydrogen.

You can observe intensity alternation with any method that can resolve rotational lines - absorption, emission or something more sophisticated. Oxygen is one of the extreme cases: it's nuclear spin is zero, so even rotational states in ground electronic state are missing completely. If you study excitation spectra of the Schumann-Runge band, you will find rotational lines P(1), P(3),... but not P(2) and so forth. I believe that this effect was important in the early days of spectroscopy, when spectral resolution of instruments was lower, so with every other line missing one could resolve spectra much better. See "Spectra of diatomic molecules" by G. Herzberg.

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  • $\begingroup$ Thanks for your post, but could you help me understand what this alternating intensity effect actually is observationally, what the multipolarities are, etc.? $\endgroup$ – user4552 Sep 8 '13 at 20:24
  • $\begingroup$ I am not sure what you mean by is. What you have is an uneven mixture of two isomers of a molecule. Any observation that can distinguish rotational states will show different intensity for even and odd lines because the amount of molecules is different. And sorry, I don't know what multipolarity is. $\endgroup$ – gigacyan Sep 10 '13 at 20:20

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