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I read in several papers that DFT is notoriously bad at describing free oxygen molecules but in none of them an explanation or reference concerning the problem has been provided. I guess this has something to do with its triplet ground state but I haven't come across a good explanation yet. Could you help me out?

Update: If someone wants to know the papers I'm refering to, here are the links and the passages that are most relevant (both groups use gradient-corrected (GGA) density functionals, so maybe this isn't a general problem of DFT but only of GGA- or LDA-functionals):

  • Rossmeisl et al. have devised a method for calculating the Gibbs Free Energy of the intermediate in the oxygen evolution reaction. The intermediates are treated with DFT but in order to calculate the Free Energy of the final products, including desorbed oxygen molecules, they use tabulated values.

    The free energy change of total reaction: $\ce{H2O -> 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.

    There are other groups studying water splitting that use the experimental value too, instead of calculating it with DFT.

  • The paper by Bloechl includes calculations of the binding energies of a hydrogen molecule, which are in rather good agreement with experiment (4.338 eV (DFT) vs. 4.448 eV (exp.)), and of an oxygen molecule, where the DFT value is off by 0.795 eV.

    The binding energy of an oxygen molecule is calculated to be 5.912 eV, including the zero-point vibartion energy of 0.110 eV, which is comparable to the atomization energy of 5.906 eV, obtained with the Becke-Perdew-Wang gradient-corrected density functional. At 5.116 eV, the experimental binding energy is substantially smaller (by 0.795 eV) than the theoretical prediction.

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  • $\begingroup$ @Philipp So, off the cuff, DFT in general notoriously bad at describing open-shell systems. Molecular oxygen may have an even number of electrons but the ground state of O$_2$ is a triplet state where two electrons are unpaired. It would be my guess that perhaps a problem lies with the exchange portion of a DFT method. Its tough to say (given the speedy development of functionals these days) if any functional exists that would properly characterize O$_2$. However, many commonly used functionals are expected to fail in this type of situation. $\endgroup$ – LordStryker Oct 31 '13 at 19:08
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The answer is most certainly the Self interaction error of an electron with itself in the exchange term.

This would probably be more helpful than what I can write here, as I have little time to make this answer more complete.

A. Droghetti PRB 2008

Also have a look at Klupfel2012

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  • $\begingroup$ Thanks for providing the links and the keyword "Self interaction error". $\endgroup$ – Philipp Nov 4 '13 at 19:17
  • $\begingroup$ No problem, if you are looking for keywords, see also Self-Interaction Correction (SIC). I think that the first serious work on this (with relation to what you are talking about) was here: http://dx.doi.org/10.1063/1.476859 $\endgroup$ – Dr_bitz Nov 6 '13 at 14:35
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I would make sure to compare the correct endpoints of the "reaction". The molecule should be a triplet (i.e. spin-polarized) and the separated atoms should be spin-polarized and Hund-compliant. Similarly, the expt energy itself should be the atomization energy of the triplet molecule into spin-polarized atoms.

That said, a typical value for the binding energy of the triplet (without the zero point vibration quantum, which reduces it) is about 4.95 eV/atom in GGA compared to spin-unpolarized atoms.

On the SIC, the first serious SIC work was actually Phys Rev B 23, 5048 (1981). A good practical SIC, unfortunately used less than it deserves, is Phys Rev B 84, 195127 (2011), also reviewed in a previous version in Eur. Phys. J. B 71, 139 (2009).

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  • $\begingroup$ You can add links to the references you used to appreciate them. $\endgroup$ – Nisarg Bhavsar May 4 at 6:02

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