Why is it so difficult to perform DFT calculations that consider temperature?

I have seen that time-dependent DFT is needed to model systems at high temperature. Why is this the case? What about finite temperature functionals? Why is it also acceptable to use zero-temperature functional with TD-DFT?

I have also read that thermal fluctuations can result in the occupation of electronic states (excited states) above the ground state.

I have tried to rationalise the fact that the electron density changes with temperature as:

If the ions in the until cell move, their degree of orbital overlap and electron density will change. The degree of orbital overlap influences the band-structure, which then impacts all of the electronic and physical properties.

Is this accurate?


2 Answers 2


One thing I will request you to note here is the definition of temperature. Temperature, in broad sense, is defined as the average kinetic energy (for equilibrium conditions, I hope you are speaking of that) of a system. The contribution mostly arises from translational kinetic energy, and the fluctuation in temperature is proportional to $\frac{1}{N-1}$. (I don't remember if it has some power or something.) In other words, a particle has infinite temperature fluctuation. So if you want to do finite temperature DFT calculations, you will need an ensemble of particles, and it will only add the cost.

In case you have infinite computational power (or at least enough to support no of molecules in the order of Avogadro number), you can, in principle, go for DFT at a finite temperature.

If you are worried about occupation of the higher states, you can calculate the higher states using Kohn-Sham Hamiltionian (these are just eigenfunctions of the Hamiltionian), and then calculate the occupation at a temperature using the formula for the concerned ensemble. Since the states of the KS hamiltionian are its eigenstates, these should not change if you have a static distribution of electrons. The eigenstates mean that these are decoupled by the same KS Hamiltonian, and in principle, you are not allowed to take an electon from one state to other without external purturbation that couples the states.

However, at a given temperature, everything is dynamic. This invokes that the states have finite lifetime, which from Uncertainty principle, imposes broadening in energy, i.e. it is not an eigenstate anymore. This imply there can be coupling between different states, and hence electrons can jump between states without external perturbation. You can externally add broadening to the states using broadning matrix, that mimics the temperature to some extent.

For TDDFT, you actually introduce an alternate electric field, and then this purturbation is used to calculate coupling between the concerned states. Thus, TDDFT calculates the excitation using light. It has nothing to do with the temperature.

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    $\begingroup$ Thank you for your great explanation. Can you please explain how in this paper (given below) has calculated the overall energy barrier at two different temperatures (0 K and 298 K). They explained "The overall energy barrier at 0 K for the LH and ER mechanisms is 0.63 and 0.77 eV, respectively. At 298.15 K, the overall energy barrier for the LH mechanism decreases to 0.58 eV, whereas that for the ER mechanism increases to 0.88 eV, which implies that CO oxidation on Ni-Gr prefers to proceed via the LH mechanism kinetically." DOI: 10.1039/C6NJ00924G $\endgroup$ Sep 6, 2020 at 20:27

Temperature is difficult to capture in any electronic structure theory (not just DFT) because almost all of them invoke the Born-Oppenheimer approximation. The BO approximation assumes the electrons instantaneously equilibrate to the current nuclear configuration (geometry). The difficulty arises then because most theories assume the nuclei are stationary as they solve for the electronic structure (either the wavefunction, or in the case of DFT, the density). As you cannot clearly define temperature without nuclear motion, these methods cannot capture temperature. As Mitradip notes, you would need to use ensemble or finite-temperature DFT to recover a description of temperature. AFAIK, you cannot use just TD-DFT to capture thermal effects. You need to recover a thermal distribution (which the ensemble achieves).

As for the thermal excitation of electronic states: You would only get electronic excitation if the thermal energy was enough to surpass the energy of an excited state. Unless you are at very high temperatures or have systems that have low-lying electronic excited states, it's not so much of a concern as is the excitation of vibrational (and of course rotational) states.

However, if you do have significant thermally induced electronic excitation, then inclusion of thermal effects would certainly affect the predicted observables of your material under study.

  • $\begingroup$ Is it possible to calculate the total energy of a system at a finite temperature using vasp? Does VASP implement finite temperature calculations? $\endgroup$ Sep 16, 2023 at 22:37

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