# Given that Kohn-Sham DFT is strictly a ground-state method (at 0 K), how is it sufficient to describe materials in real-life applications?

Kohn-Sham DFT appears to be so popular even though it is strictly a ground-state method - all calculations are done at 0 K. How then, is it so popular when describing materials that have real-life applications (possibly at room temperatures, or much higher temperatures)? Or am I mistakenly deeming KS-DFT to be too popular when there are other more suitable methods like TD-DFT, GW-BSE etc?

• +1. It's a good question. You are not mistakenly deeming KS-DFT to be more popular than TD-DFT and GW-BSE. KS-DFT is undoubtedly several orders of magnitude more popular, probably because it's cheaper and not much less accurate? I mean there's so many approximations in all of those three methods anyway, so might as well use the cheapest one for most applications. I would answer if I had specific examples I could give of KS-DFT being sufficient for something, or of KS-DFT and TD-DFT/GW-BSE being equally clueless of the true answer for something. – Nike Dattani Jun 9 '20 at 3:48

These are a few extra points to complement Andrew Rosen's comprehensive response:

1. To be absolutely clear, typical DFT calculations are not performed at 0K, a better description of what happens is that they are performed "for a static crystal". Static crystal means that the atoms are fixed at their crystallographic positions (which is what a typical DFT calculation does), but this situation is different from 0K. Even at 0K, the atomic nuclei move due to quantum zero-point motion (also called quantum fluctuations). This quantum zero-point motion is very small for compounds containing heavy elements, which means that in these cases "0K" and "static crystal" are almost the same and many people use the terms interchangeably. However, quantum zero-point motion can be very large for light elements, for example in solid hydrogen (the lightest of all compounds) it dominates even at room temperature. To perform a calculation at 0K (as opposed to a static crystal calculation), the strategy that Andrew Rosen proposed would still work if you use quantum statistics. For example, solving the vibrational problem within the harmonic approximation at the quantum mechanical level with give you the zero-point energy as the ground state of a quantum harmonic oscillator. A good reference for hydrogen is this review article.

2. Although Andrew Rosen is correct in that the potential energy surface is largely temperature independent, there are a few situations in which the potential energy surface does vary significantly with temperature. An example of this is provided by many perovskite crystals (e.g. $$\ce{BaTiO_3}$$), which undergo a series of temperature-induced structural phase transitions. For example, at very high temperatures the structure is cubic, and this structure sits at a minimum of the potential free energy surface. However, if you calculated the potential energy surface (instead of the free energy surface), then it has a double-well shape and the cubic structure is at the saddle point, so the structure will lower its energy by distorting and going to one of the minima. This is precisely what happens at low temperatures because the corresponding free energy surface acquires a double-well shape, and the system undergoes a structural phase transition. A good early DFT reference for this is this paper.

3. As already pointed out by others, the differences between DFT and TDDFT or $$GW$$-BSE are not related to the inclusion or exclusion of temperature. In this language, both TDDFT and $$GW$$-BSE have no temperature for the behaviour of the nuclei, and you could incoporate it in a similar fashion to what Andrew Rosen described for DFT.

• Many thanks for these important clarifications! Absolutely agree on all fronts. – Andrew Rosen Jun 9 '20 at 14:15
• Would you have an issue with saying DFT is performed at 0k under the Born-Oppenheimer approximation? I guess I am questioning the need to bring in crystals. Nuclei positions don't move, but that is due to the BO approximation, not the need to solve a crystal... or am I missing something? – B. Kelly Jul 22 '20 at 17:57
• @CharlieCrown, your question could be discussed over many lines, and terms are certainly abused in the literature. Let me first clarify that all along I am talking about lattice temperature (as opposed to electronic temperature). For lattice temperature, then DFT does not correspond to 0K: the BO approximation leads to two equations, the one for the electrons (that DFT solves) but also one for the nuclei. You need to solve the one for the nuclei at 0K to obtain the 0K result. Many people ignore this and say that DFT is 0K, but what they should say is that DFT is "static". – ProfM Jul 22 '20 at 18:10
• @CharlieCrown and the argument is the same for crystals or molecules, if the nuclei are completely fixed, then no lattice temperature or zero-point quantum motion is included. – ProfM Jul 22 '20 at 18:12
• @ProfM Okay, I follow now, thanks for the clarification :) – B. Kelly Jul 22 '20 at 18:32

You are correct that KS-DFT, strictly speaking, involves calculations of a potential energy surface at 0 K. However, if you accept that the density functional approximation you are using is sufficiently accurate, it is not too difficult of a stretch to go from 0 K to finite temperature conditions for an application of interest. The key assumption is that the potential energy surface itself does not significantly change from 0 K to finite temperature. Admittedly, I don't know of many examples of molecules or materials where this has been shown not to hold, but there are surely some I'm not aware of. As a side-note, I have read that for very hot, dense matter, there is a need to account for thermal (i.e. entropic) effects on the exchange-correlation energy, and this has led to some developments in thermal DFT (e.g. as discussed here). Otherwise, is a very common assumption that the 0 K potential energy surface is relatively unchanged.

With knowledge of the 0 K structures and vibrational modes, you have essentially everything you need to compute thermochemical quantities at finite temperature. It's simply a matter of computing the relevant translational, rotational, vibrational, and electronic partition functions. There are several assumptions that go into each one of these components (e.g. ideal gas approximation, rigid rotor approximation, particle-in-a-box approximation, harmonic oscillator approximation), but that has relatively little to do with the fact that KS-DFT is at 0 K and more that you need some way to express the partition functions. There are countless references on this topic to choose from. For molecular systems, Chris Cramer's "Essentials of Computational Chemistry: Theories and Models" does a great job. For periodic DFT calculations, you may find the thermochemistry summary on the ASE webpage to be helpful. I have also put together a series of notes covering this topic on my website Rosen Review.

Also, it's not as if TD-DFT isn't widely used. It's the method of choice for computing excited states, which could be useful for predicting UV-Vis or X-ray absorption spectra. To the best of my knowledge, the main use-case for TD-DFT is because KS-DFT is a ground state theory, not because KS-DFT is strictly true at 0 K.

1. Rather than deriving the orbital equations from a minimization of the energy, $$E$$, one minimizes the free energy $$F = E - TS$$, where $$S$$ is the entropy. A practical consequence is that one needs approximate kinetic, exchange, and correlation functionals that are functionals of both the density and the temperature. There are not very many of these out there, and they are mainly used in high-density plasma physics contexts, rather than conventional materials modeling.
2. When forming the density from the spin-orbitals, one must use proper finite-temeprature occupations numbers, rather than just summing states whose energy is less than the chemical potential (Fermi level). In practice, instead of evaluating the density as $$\rho(\vec r) = \sum_{\epsilon < \mu} |\psi(\vec r; \epsilon)|^2$$ (where $$\psi(\vec r; \epsilon)$$ is the spin-orbital with eigenvalue $$\epsilon$$), one should instead evaluate $$\rho(\vec r) = \sum_\epsilon n(\epsilon) |\psi(\vec r; \epsilon)|^2$$, where the sum runs over all eigenvalues and $$n(\epsilon) = \frac{1}{1 + e^{(\epsilon-\mu)/kT}}$$ is the Fermi-Dirac occupation number. At zero temperature, $$n(\epsilon)$$ becomes a step function at $$\epsilon=\mu$$, and the usual density expression is recovered.