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I am working in my final undergraduate Project about predicting belite (Ca2Si04) polymorphs using DFT calculations and I need to justify why my SIESTA run is looking for minimum enthalpy and not free energy. I have a vague idea that is because DFT is perfomed using T=0K but I cannot find any article that supports it. I need to use references, so please if you could provide it I'd grateful.

Thanks!

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Jan 12 at 9:48
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    $\begingroup$ $G = H - TS$. What happens when $T=0$? $\endgroup$
    – Hayden S
    Jan 12 at 22:19
  • $\begingroup$ Do you actually aim at predicting polymorphs at zero temperature? If yes, then Hayden's answer (H - 0 * S = H) works perfectly well. If no, then the answer will involve other factors like, e.g. it is much more difficult to minimize the free energy than minimize the (zero-temperature) enthalpy. $\endgroup$
    – wzkchem5
    Jan 13 at 12:54
  • $\begingroup$ I found another similar question here : mattermodeling.stackexchange.com/questions/1228/… My calculation is not done at zero temperature (at least I haven't specified it) but I think that DFT itself does it because the Born-Oppenheimer approximation makes the nucleus static so temperature makes no sense... Well, that is what I think, and if that is true I need references $\endgroup$
    – leire
    Jan 13 at 14:34
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    $\begingroup$ Here is a widely cited go-to reference that works 0K and >0K calculations and phase diagrams: journals.aps.org/prb/abstract/10.1103/PhysRevB.65.035406 There are many more similar ones by Reuter around the same time (2001-2003 or so). And a book chapter: link.springer.com/chapter/10.1007/978-1-4020-3286-8_10 They focus on surfaces, but the same workflow applies to polymorphs and everything else. $\endgroup$
    – Andrey Poletayev
    Jan 16 at 20:55

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DFT is most commonly used to calculate minimized energy structures. This is likely what your project is on, unless you're doing ab initio molecular dynamics, or excited state DFT (e.g. time-dependent DFT) calculation. In this case, your calculations are equivalent to finding the lowest enthalpy ground state minus the vibrational zero point energy. Enthalpy $H$ will be equal to the total energy $E$ unless you're also applying some external stress (i.e. pressure), which would correct the enthalpy by a $PV$ term.

Since your calculations are at $T =0$ (equations of motion are not even employed), the energy (or enthalpy) minimization is also equal to the Gibbs free energy $G$.

If you want to determine which polymorph is most stable at finite temperature and pressure (lowest $G$), you need a more sophisticated calculation than straightforward DFT structure minimization. A common method is to correct the minimized energy for vibrational energy based on a harmonic or quasiharmonic vibrational calculation.

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  • $\begingroup$ In equilibrium, the magnitude that is minimized is the total energy I guess but in many articles I find that what is minimized is the free energy. From G they derive that at T=0, G is equal to H and finally if no pressure is applied, G=H=U. You say that want needs to be minimized is H and then derive that it is equal to G...I find that a bit confusing $\endgroup$
    – leire
    Jan 22 at 12:51
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    $\begingroup$ I wanted to clarify the T=0 thing. I have read that as ordinary DFT is performed for static crystals, no temperature exists for something that is static. Real structures do have movement so DFT is not completely suitable for describing them, however, DFT is used because it is a good approximation. Even at T=0 structures have the zero point energy (=a bit of movement) so it is not a real T=0 calculation either. This zero point energy can be quite small (especially for heavy elements) so it is almost accurate to say that DFT is for structures at T=0. Is this correct? thanks $\endgroup$
    – leire
    Jan 22 at 13:00
  • $\begingroup$ When I say ordinary DFT I mean the Kohn Sham one. I am a beginner here who is still understanding the most basic things :) $\endgroup$
    – leire
    Jan 22 at 13:04
  • $\begingroup$ In my first DFT calculation I am not aplying pressure but in the next one I may need to... Is the ordinary DFT a correct method to do so? $\endgroup$
    – leire
    Jan 22 at 13:07

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