8
$\begingroup$

One of the challenges of modelling completely new material using DFT is the selection of initial geometry. Usually, an experimental structure (cif) which appears to be similar to the new material in question is selected intuitively (correct me if I'm wrong) and several such selections are made to comparatively analyse the optimised free-energies if necessary. I found a similar question that concludes optimised geometry depends heavily on the initial geometry.

Let us assume the full relaxation of ionic positions, cell volume and cell shape is carried out starting from some geometry.

My question is, is there a possibility to change the initial crystal system during relaxation? Can it change from orthorhombic to cubic, tetragonal etc., for example, in the process of finding a local minimum in the potential energy surface?

If it does not change even after the relaxation, does that mean the selected initial geometry is a good starting point?

$\endgroup$
  • 1
    $\begingroup$ Can you explain how this is different than the question you linked to? Also, you refer to finding a saddle point on the potential energy surface, but presumably you mean a local minimum? $\endgroup$ – Andrew Rosen Aug 9 at 14:58
  • $\begingroup$ Here what I am mostly interested in is, given the permission and, no symmetry is assumed in VASP INCAR (ISYM=0 or -1), what degree of structural change is possible during a relaxation. For example, can an orthorhombic cell relax to a much higher symmetry such as cubic? And yes, I meant a local minimum, but in this case the program should be able to find a lower local minimum because the degrees of freedom is higher than that in a normal ionic relaxation preserving the cell shape and possibly the symmetry. $\endgroup$ – Achintha Ihalage Aug 9 at 15:51
7
$\begingroup$

In general this depends on the potential energy surface. If symmetry is turned off and the cell is allowed to change shape and size, then no promises are made either way on what will be found. Shallow local minima may be skipped over due to the optimization process, which will likely send you to a higher symmetry if that is the most stable structure.

This is easy to imagine, take an FCC cell (of something that actually prefers FCC like Pt) and displace one atom 0.2A or so. You can also extend one cell vector by a similar amount. Run with your given settings and watch as the cell comes back to the original FCC cell with its symmetry. This is a bit of a contrived example, but it shows that you can go from one symmetry to another via relaxation.

Going from high symmetry to low symmetry typically causes problems if symmetry is turned on. This often manifests itself as an error in the software though. A good example of high symmetry to lower symmetry is an adsorbate moving from a top site to a bridge site on a Pt(111) surface. This can occur if the top site is a local maximum rather than a local minimum. You can see below that this would lower the symmetry of the cell. This is a quite common problem to see.

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the explanation. Could you tell me if you have observed the symmetry increasing phenomenon in the example FCC-Pt yourself? If so, had you turned the symmetry on or off in the calculation? $\endgroup$ – Achintha Ihalage Aug 9 at 17:45
  • 1
    $\begingroup$ symmetry off with a rattled cell (random displacements of about 0.1A) and an extended X vector. I ran using GPAW but VASP should be the same. $\endgroup$ – Tristan Maxson Aug 9 at 19:26
  • 1
    $\begingroup$ Good answer. It should also be mentioned that it will strongly be dependent on the optimization algorithm as well. $\endgroup$ – Andrew Rosen Aug 9 at 19:50
  • $\begingroup$ I think as long as the optimization algorithm doesn't attempt to use symmetry that the calculator isn't aware of, it should be mostly independent of optimizer. $\endgroup$ – Tristan Maxson Aug 9 at 19:58
4
$\begingroup$

A geometry relaxation is a calculation that locates a local minimum of the potential energy surface. Symmetry can change during a geometry relaxation, but the interplay between symmetry and minima of the potential energy surface is interesting. If you enforce the symmetry of the starting structure during your geometry relaxation, then you encounter two situations:

  1. From low to high symmetry. Imposing a symmetry to your structure means that a set of symmetry operations are definitely obeyed by that structure. However, this does not forbid additional symmetry operations to also be obeyed by that structure, and therefore it is possible to go from low to high symmetry. The higher symmetry is only enforced numerically, so it is advisable to impose the higher symmetry after the geometry relaxation to include the newly found symmetry operations exactly. Tristan Maxson provided a nice example of this in their answer. As another example, take diamond in the cubic fcc structure, reduce the symmetry to tetragonal to create your initial structure, and then when you relax the structure it will fall back to the cubic structure increasing its symmetry.
  2. From high to low symmetry. If you are enforcing a certain symmetry in your structure, but the real structure you are looking for has a lower symmetry, then it is impossible to find it during your geometry relaxation. This is because imposing a symmetry on your structure effectively reduces the dimensionality of the potential energy surface by removing all "directions" that are not consistent with the symmetry you are imposing. This means that when you relax imposing symmetry you will fall to a local minimum of this symmetry-constrained lower-dimensional energy surface, rather than to a local minimum of the real energy surface. If the real ground state of the system has a lower symmetry, then this means that the minimum in the symmetry-constrained lower-dimensional energy surface is in reality a saddle point in the full energy surface, but you have no way of finding this out with your symmetry-constrained geometry relaxation. However, a good way to test for this once you have found your structure with a symmetry constraint is to perform a subsequent phonon calculation. A phonon calculation gives you the Hessian about the stationary point in the potential energy surface, and if you are at a saddle point of the full energy surface this will come out as a negative eigenvalue of the Hessian, corresponding to an imaginary phonon frequency. The associated eigenvector tells you in which direction to distort the structure to find the lower energy lower symmetry structure.

If you do not enforce the symmetry of the starting structure, then in principle you can change from low to high symmetry or from high to low symmetry during the relaxation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Many thanks for this informative answer. Regarding the statement, "if the initial geometry does not change even after the relaxation, that means this initial geometry is a good starting point", do you have any opinion on this? Is it generally correct? $\endgroup$ – Achintha Ihalage Aug 10 at 10:11
  • 2
    $\begingroup$ @AchinthaIhalage this will depend on whether you impose the initial symmetry or not. If you impose it, then this corresponds to my part of the answer "from high to low symmetry": you cannot know by only doing the relaxation. My suggestion would be, as Tristan Maxson said in one of their comments, to "shake" the structure a little (e.g. displacing atoms by a small random amount) and then see if it goes back to that structure. Another would be to do a phonon calculation after the geometry optimization to see if you find any imaginary modes. $\endgroup$ – ProfM Aug 10 at 10:19
  • $\begingroup$ I think there are two different core questions. Symmetry can prevent you from finding the right geometry technically, but not enforcing symmetry does not ensure you will find the right geometry. That requires phonon / MD. $\endgroup$ – Tristan Maxson Aug 10 at 16:23
  • $\begingroup$ @TristanMaxson absolutely agree with this, and I would add that it may be even necessary to go beyond phonon/MD to find the right geometry: one may have to do full-fledged structure prediction. $\endgroup$ – ProfM Aug 10 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.