I want to know which particular calculations are suitable for each of them. Is SIESTA free, as in can you work on it without license. (question is based on the fact for now that all these will be done locally at home and not at institute where such software’s are available) If there is some other options available, please let me know.

  • 4
    $\begingroup$ An overall comparison between Quantum Espresso and SIESTA may be too broad. It would help to know what sort of calculations you are hoping to do and then users could point to the pros/cons of each program for those application. There may be a number of other options available that can do what you want, but its difficult to say without at least a general idea of the types of calculations you are interested in. $\endgroup$
    – Tyberius
    Commented Jul 13, 2021 at 18:30

1 Answer 1


Comparing what each software is capable off, is relative easy: open both webpages and look for feature pages.

From SIESTA project page:

  • Total and partial energies.
  • Atomic forces.
  • Stress tensor.
  • Electric dipole moment.
  • Atomic, orbital and bond populations (Mulliken).
  • Electron density.
  • Geometry relaxation, fixed or variable cell.
  • Constant-temperature molecular dynamics (Nose thermostat).
  • Variable cell dynamics (Parrinello-Rahman).
  • Spin polarized calculations (collinear or not).
  • k-sampling of the Brillouin zone.
  • Local and orbital-projected density of states.
  • COOP and COHP curves for chemical bonding analysis.
  • Dielectric polarization.
  • Vibrations (phonons).
  • Band structure.
  • Quantum Transport.

From Quantum-ESPRESSO (QE) page:

  • Ground-state calculations.
  • Structural Optimization, molecular dynamics, potential energy surfaces.
  • Electrochemistry and special boundary conditions.
  • Response properties (DFPT).
  • Spectroscopic properties.
  • Quantum Transport.

Personal opinion. I tested several DFT software for periodical systems. Today, I only use SIESTA and can run simulations in regular desktop (8 cores, 16GB RAM) and even in my notebook (12 cores, 8GB RAM). Recently we run a calculation in QE that failed after one week (in a workstation with 64GB RAM and 32 cores). The same calculation run smoothly in my PC in 30 minutes (8 cores, 16GB RAM).

One of the main complain about QE is that it needs huge computational resources: too much RAM and if you don't have enough RAM, too much hard disk space.

Taking a careful read about what each software can do, you will find that SIESTA is a little limited compare to QE.

About the license type:

  • 3
    $\begingroup$ Perhaps you could add that since v. 4 Siesta is released under GPL making it free to use. That is partly a question from the OP. $\endgroup$
    – nickpapior
    Commented Jul 14, 2021 at 8:29
  • 2
    $\begingroup$ "Recently we run a calculation in QE that failed after one week (in a workstation with 64GB RAM and 32 cores). The same calculation run smoothly in my PC in 30 minutes (8 cores, 16GB RAM)." Surely these are not the same calculation since QE uses plane waves whereas SIESTA uses pseudoatomic orbitals. Using an atomic orbital basis allows you to get semiquantitative results very quickly, but nailing down the converged value may become difficult. With plane waves this is easier, but it may be computationally intractable. Both, of course, suffer from having to use pseudopotentials. $\endgroup$ Commented Jul 14, 2021 at 19:26
  • $\begingroup$ @SusiLehtola the same calculation refers to the type of calculation like geometry optimization, single point, for example. Not to the method itself, that it is, of course, different. $\endgroup$
    – Camps
    Commented Jul 14, 2021 at 19:44
  • 1
    $\begingroup$ @Camps but they are fundamentally different. Even within the LCAO sense a minimal-basis calculation differs from a quadruple-zeta calculation. Sure you can probably run the former for a huge system, but there's no guarantee that the results will make sense. $\endgroup$ Commented Jul 17, 2021 at 16:38
  • 2
    $\begingroup$ Thank you sir for this detailed answer. $\endgroup$ Commented Jul 19, 2021 at 11:32

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