When 2D compounds have a dipole, the potential distribution becomes linear (the left panel of the figure). After considering the dipole correction, the potential has two plateau regions in the vacuum (the right panel of the figure), rather than one. How can we determine the vacuum potential from the potential distribution curve? Physically speaking, we can either use the right- or left-side vacuum potential, or average them?

point.1, 2, 47 & 48 are different configurations of the same material with some thermal distortions

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    $\begingroup$ Looking at your plots, it seems like the electric field $E_z(x,y)$ reverses sign depending on the point (x,y). Is this what you expect? $\endgroup$ Jan 10, 2022 at 21:49
  • $\begingroup$ Thank you Leopold! I actually generated four different configurations, named as point.1, point.2, point.47, and point.48. These configurations correspond to four different types of lattice distortions at finite temepratures. But I am not sure how to align the vacuum level to get an thermal average of the VBM and CBM. Do you have any ideas? $\endgroup$
    – Bo Peng
    Jan 14, 2022 at 11:47
  • $\begingroup$ I see, so just to clarify: the four different points refer to the x-y average of the electrostatic potential for four different snapshots of the 2d layer? Is it expected that these thermal vibrations change the polarization direction of the layer? $\endgroup$ Jan 14, 2022 at 18:45
  • $\begingroup$ Yes, these are in-plane x-y average of the electrostatic potential for four different snapshots of a 2D layer. I think it is possible that the thermal vibrations change the polarization direction of the layer. And point 1 and point 2 (or point 47 and point 48) are opposite vibrational patterns. $\endgroup$
    – Bo Peng
    Jan 16, 2022 at 0:04

1 Answer 1


The purpose of the dipole correction is to remove the spurious electric field that arises between the polarized 2d layer and its periodic replica (think of the correction as if it made your calculation non-periodic along z).

However, there is still a physical, finite electric field inside your layer, pointing from the positively charged side to the negatively charged one. When moving a test charge through your layer, that electric field will do work on the test charge, and thus the vacuum potential [1] (=the potential "just outside the material") is different between the top and the bottom of the layer.

Now, if I understand your plots and comments correctly, in this particular material the direction of the polarization is not static: at finite temperature, even the orientation of the polarization swings back and forth over time. There are two possibilities: (a) over time, the vacuum potential on the top and the bottom of the slab average out to the same value (polarization is washed out by thermal vibrations), or (b) a finite difference remains between the average vacuum potential at the top and the bottom.

If (b) is the case, then, in principle, it matters whether you are interested in the vacuum potential at the top or the bottom of the layer.

Note, however, that you are simulating a 2d material in vacuum, which does not exist in the real world. In reality, your material will be supported by a substrate, which can screen the electric field, modify the vibrations, etc.

[1] In case it gives you some comfort, I also struggled with the term "vacuum potential" (aka "vacuum level") in the past, since I would intuitively associate it with the energy of an electron (or positron, whatever sign you prefer) at rest, infinitely far away from the material. That energy (let's call it the "infinity level") is of course always the same by definition.

How does one reconcile this with the fact that, in your calculation, the potential at the top and the bottom of the layer will always be different, no matter how far away you go? The issue here is that your layer's infinite size along x and y makes it impossible to get away from it. Once you make your layer finite (non-periodic) in x,y it will give rise to weak, long-range electric fields (at the length scale of the dimensions of the layer), which will align the "infinity level" on both sides of the layer. A difference in the vacuum level on both sides of the slab, however, would remain.

  • $\begingroup$ Great answer! :) $\endgroup$
    – nickpapior
    Jan 27, 2022 at 9:06
  • $\begingroup$ Thank you Leopold! This is super helpful!!! I will think about this a little more and get it back to you asap. Thanks once again! $\endgroup$
    – Bo Peng
    Feb 23, 2022 at 9:49

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