Although the ferroelectric polarization is usually calculated with primitive cell using the Berry phase method, supercells are sometimes needed for particular applications, e.g., defects, doping, etc. As polarization is defined as the dipole moment divided by the volume, using larger supercell would not change the polarization in principle. However, due to the multi-value nature of the polarization according to the modern theory of polarization, the polarization of crystals is only defined modulo the polarization quanta $$\mathbf{P} = \mathbf{P}_0 + \sum_{i=a,b,c} n_i \frac{e\mathbf{R}_i}{\Omega}$$ where $n_i$ is an integer, $\mathbf{R}_i$ is the lattice vector, and $\Omega$ the volume. As can be seen, supercell would change the polarization quanta ${e\mathbf{R}_i}/{\Omega}$, thus change the calculated polarization $\mathbf{P}_0$. So I'm wondering if there is any recipe or remedy on the polarization calculations with supercells.


1 Answer 1


I'm not sure if I understand your question completely, especially this part:

As can be seen, supercell would change the polarization quanta $e\textbf{R}i/\Omega$, thus change the calculated polarization $\textbf{P}_0$.

I'm guessing $\textbf{P}_0$ here stands for the polarization modulo $e\textbf{R}i/\Omega$. If that's the case, then I don't think using a supercell instead of a primitive cell will change the value of $\textbf{P}_0$.

The macroscopic polarization in the framework of the "modern theory of polarization" is calculated by taking the difference of values calculated with centrosymmetric structure and ferroelectric structure, assuming they are on the same "branch".

Using a supercell can indeed change the polarization quantum, but these changes take place in both centrosymmetric and ferroelectric structures and because the polarization of these two structures needs to be on the same "branch", they have the same changes. In the end, the effect of using a supercell is canceled out.

I hope this is helpful =)

  • $\begingroup$ +1 Not a bad first answer. Welcome to our community! Thanks for your contribution and we hope to see much more of you!!! Hopefully the user that asked the question can clarify what that sentence means. $\endgroup$ Dec 29, 2020 at 18:26
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    $\begingroup$ I think you are right. If the polarization is calculated along the branch, the changes should keep same. $\endgroup$ Dec 29, 2020 at 19:18

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