I have a material that can crystallize in either a cubic or tetragonal polymorph depending on the synthesis. I used DFPT in VASP and calculated the phonon band structures. I see that the cubic has all positive frequencies, while the tetragonal has some negative (imaginary) frequencies. This leads me to believe that the cubic phase is at a local minimum on the potential energy surface, while the tetragonal is at a saddle point.

I thought the tetragonal might be stabilized by entropic contributions at finite temperature, so I used phonopy to plot the thermal properties. I was looking at the Helmholtz free energy vs temperature, and surprisingly at 0 K, the free energy of the tetragonal phase is slightly lower than the cubic. Once the temperature increases, a crossover happens, and the free energy of the cubic becomes lower.

My question is: why is the free energy of the tetragonal phase at 0 K lower than the cubic, when the tetragonal phase has negative phonon frequencies and the cubic does not?

  • $\begingroup$ Why a saddle point can't have lower energy in one system than the local (global) minimum of the other system? $\endgroup$
    – Camps
    Feb 22, 2021 at 11:25
  • $\begingroup$ They are the same composition (or system). Just two different polymorphs that depend on the synthesis conditions. $\endgroup$
    – DoubleKx
    Feb 22, 2021 at 21:56

1 Answer 1


Your interpretation of the results. I agree with you that if you find no imaginary frequencies in the cubic phase it means it is at a local minimum of the potential energy landscape, and that if you do find imaginary frequencies for the tetragonal phase, then that one is at a saddle point. I also agree that a phase exhibiting imaginary frequencies may be stabilized by entropic contributions (see more below). Just a small caveat at this point: entropic contributions typically favor higher symmetry phases, so I am a little surprised that you find a stable cubic phase but an unstable tetragonal phase. This suggests that the cubic phase is the low temperature phase, while the tetragonal phase is the high temperature phase. In many compounds (e.g. perovskites) this is usually the other way around. Either way, the comments below apply to any unstable phase.

Finite temperature stabilization. While you are correct that a phase exhibiting imaginary phonons may be stabilized by entropic finite temperature contributions, your approach to evaluate these looks incorrect. If I understand what you are doing, then you are evaluating the Helmholtz free energy as a function of temperature using the harmonic approximation. While this is fine for a phase without imaginary phonons, this is incorrect for a phase with imaginary phonons. How would you include the contribution of the imaginary modes to the free energy? Instead, what you need to do is to include anharmonic terms in the expansion of the potential energy landscape. This is the only way in which you may be able to accurately evaluate the Helmholtz free energy of a phase that exhibits imaginary phonons at the harmonic level.

Anharmonic calculations. There are a variety of strategies to include anharmonic terms in the calculations to evaluate the Helmholtz free energy. Methods range from molecular dynamics to the self-consistent harmonic approximation, and which one is more appropriate will depend on the details of your system. However, their computational expense is significantly larger than that of harmonic calculations.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .