I relaxed Na2S2 molecule in QuantumEspresso and plotted the planar average of potential energy across z axis so as to find the Vacuum potential. This is the output I got : Planar average of potential  of Na2S2 As you can see the vacuum potential is not constant making it impossible to calculate it. While following the same procedure for Na2S I got the following: Planar average of potential  of Na2S as you can see the vacuum potential is a constant value. What could be a reason for what happened in the case of Na2S2? How can I solve it ?

Thank you!

The input file for the geometry optimization of Na2S2:

ibrav = 4,
a = 14.58 ,
c= 20,
nat = 4
ntyp = 2
vdw_corr = 'dft-d3'
ecutwfc = 50.0 ,
ecutrho = 250.0 ,
electron_maxstep = 200,
conv_thr = 1.0d-10 ,
ion_dynamics = 'bfgs'

Na 22.989 Na.pbe-spnl-kjpaw_psl.1.0.0.UPF
S 32.065 S.pbe-n-kjpaw_psl.1.0.0.UPF

Na           -2.1578688813       -0.0711007913        0.0959862818
S             0.0279507567        0.7976264059        1.1138133631
S            -1.4993815540        2.3462755509        0.6717950344
Na            0.8998813708        2.9052562024       -0.0587466613

 K_POINTS automatic
2 2 1 0 0 0
  • 1
    $\begingroup$ I have no experience with QE but you might want to add a dipole correction: quantum-espresso.org/Doc/INPUT_PW.html#dipfield $\endgroup$
    – Fabian
    May 18, 2021 at 6:10
  • $\begingroup$ @Fabian Thank you sooo much. That got me somewhere to start. when I add dipole correction gave me two vacuum potentials. From reading, this usually refers to difference in work function at different sides of slabs(Hope I am right!). But I have a chose quite a large unit cell and its a small molecule, will it have a dipole interaction? $\endgroup$ May 21, 2021 at 17:45
  • 1
    $\begingroup$ I agree with @Fabian that this looks like a dipole interaction. Why not do some charge decomposition and see what the classical electrostatic field looks like? Remember that you are simulating an infinitely periodic set of molecules, so the field is not the same as from an isolated molecule. $\endgroup$ Jun 20, 2021 at 12:40

1 Answer 1


As mentioned by Phil Hasnip in the comments, placing a $\mathrm{Na_2S_2}$ molecule under periodic boundary conditions is akin to simulating a 3d array of these molecules. If the molecule has a nonzero dipole moment, this give rise to an electric (dipole) field between adjacent 2d "planes" of molecules. Unless you take care to orient the dipole of your molecule along a specific cartesian axis, it can point in any direction of the cell. I.e. the effect you are seeing along the z direction may as well exist along x and y with different magnitudes proportional to the component of the molecular dipole in that direction.

As with a parallel plate capacitor, at sufficient distance from the 2d "planes" of molecules the electric field is roughly constant, giving rise to the linear region of the potential in your plot. The purpose of the dipole correction is to place a counter-dipole of opposite magnitude at a position sufficiently far away from the molecule in order to "flatten" the potential curve. I'm not quite sure what you mean by "when I add dipole correction gave me two vacuum potentials". When done right (see e.g. this example by Christoph Wolf), you should simply obtain a flat potential.

By the way, if you are just interested in the value of the vacuum potential, you could place the molecule in the center of the cell and simply take the value of the potential in a corner of the simulation box. This neglects the polarizing effect of the electric field on the electron cloud in the molecule but should converge to the vacuum potential with increasing cell size.

Finally, you don't state the purpose of this calculation (and there may be valid reasons for you to use Quantum ESPRESSO) but if it is to compute quantities like the ionization potential or the electron affinity of a molecule, don't use a code with periodic boundary conditions. This is what quantum chemistry codes are for, and there are both great commercial and free implementations.


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