I'm calculating the vacancy formation energy of a 3x3x2 supercell of cerium oxide and need to include a Hubbard parameter for the Ce 4f state. However, computing this parameter separately for the large supercell would be very time-consuming due to its size (around 200 atoms). Therefore, I'm wondering if it's feasible to calculate the Hubbard U parameter for the smaller unit cell and use it for the supercell instead. Is this a valid approach or are there any drawbacks to this method?
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$\begingroup$ Are you calculating the Hubbard U self-consistently (e.g. through a linear-response formalism) or emperically through fitting of some known experimental proxy? $\endgroup$– ShaCommented Apr 10, 2023 at 9:58
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$\begingroup$ I am calculating it through linear-response, with hp.x code if Quantum Espresso. $\endgroup$– Marko_BCommented Apr 10, 2023 at 15:04
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3 Answers
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The linear-response formalism implemented in hp.x code does not require the use of computationally expensive supercells of the traditional linear-response approach to compute the Hubbard U. Actually one of the main advantages of hp.x code in Quantum ESPRESSO is the ability to determine the Hubbard U using a primitive unit cell instead of a supercell.
So in short, not just you can use a unit cell for calculating the Hubbard U but better you can use the primitive unit cell!Then use this U value for your supercell calculations.
I suggest you read the hp.x paper by Timrov, Marzari and Cococcioni (2022), which beautifully explains all this and more.
The paper is open access: https://doi.org/10.1016/j.cpc.2022.108455
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$\begingroup$ Thank you for your answer. Does this holds if I create vacancy in supercell? Can a Hubbard parameter calculated with hp.x for primitive unit cell be used for a supercell, even when there is a vacancy in the supercell? $\endgroup$– Marko_BCommented Apr 11, 2023 at 17:19
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1$\begingroup$ The short answer is NO. Have you read the paper? Please have a look at section 5 of the paper. The developers of the HP code themselves have authored several papers on its use and application, please try to learn from them. Nothing is better than learning from a firsthand account. I myself did not use the HP code, but I admire the work and research of professor Marzari and his colleagues. One will learn a lot deal from them regardless of the used DFT code. Here is another paper of them that may answer your questions: pubs.acs.org/doi/abs/10.1021/acs.jpcc.2c04767 $\endgroup$– ShaCommented Apr 11, 2023 at 18:48
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$\begingroup$ Thank you for you references.I understand that the Hubbard parameter is not transferable and is sensitive to the chemical environment. My question is about the calculation of the Hubbard U for vacancy formation energy. Can it be calculated for a smaller cell with a vacancy and then used for the supercell, or it has to be calculated directly for the supercell? Since Hubbard U will be different for systems with and without vacancies, is it legitimate to calculate the vacancy energy as the difference between these two values, considering that they were calculated using different parameters? $\endgroup$– Marko_BCommented Apr 11, 2023 at 19:45
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1$\begingroup$ As I said I did not use HP code and do not know the computational settings and details of using Quantum ESPRESSO. Please have a look at the provided papers for the computational methodology. Try to reproduce their results. The second paper includes calculating Hubbard parameters for a pristine and doped system (which is similar to your case of a defected supercell, a dopant instead of a vacancy). $\endgroup$– ShaCommented Apr 11, 2023 at 20:44
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The various methods to calculate the Hubbard U value in a first principles way don’t always give such good agreement with experimental values anyway. If it would be too expensive to calculate it with a first principles way, then you might want to just use a U value from literature, try a range of U values yourself to see the impact on the binding energy, or try a hybrid functional (although there you have to pick the fraction of exact exchange anyway).
Using a first principles method to calculate it in the unit cell is a reasonable way to try to do it, but if the supercell has a vacancy that changes the local environment, then in theory the appropriate U value could change.
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1$\begingroup$ So, if I want to calculate the Hubbard U parameter for a supercell, would the procedure be the same as for the unit cell - self-consistent calculation, Hubbard parameter calculation, variable cell relaxation calculation, and repeating until convergence? $\endgroup$– Marko_BCommented Apr 10, 2023 at 15:11
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1$\begingroup$ Yes, although I think you probably should investigate the typical accuracy of calculated Hubbard U parameters for similar systems. For one, different methods of calculating it can lead to different results, and second for some quantities the calculated ones are not good. $\endgroup$– AGSCommented Apr 12, 2023 at 6:57
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Normally this famous U parameter is a correction to reach the localized states due to the correlations which a one electron system cannot integrate. As a corrective parameter it is very variable and nobody could give the order of magnitude of its variation in a precise way because even your calculation method can affect the order of magnitude converging towards a satisfactory correction. Normally it should not vary so much, but our simulation tools are not perfect.
Concerning your question, intuitively by increasing the size of the system there is some delocalization more significant of d states than f and thus the correction is theoretically different. As stated before, nobody will be able to give the deviation rate correctly. Normally you should recalculate the parameter regardless of the size. In some cases using the same parameter can be theoretically a good approximation especially for the 4f states but I guess even more for Lutetium than for Cerium.
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1$\begingroup$ Please what do you mean by : "...... by increasing the size of the system there is some delocalization more significant of d states than f...... "? $\endgroup$– ShaCommented Apr 10, 2023 at 11:18
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$\begingroup$ This argument is relevant mainly for systems whose size varies without necessarily having periodic boundary conditions so the local electronic distribution changes and so does U as a function of size. Since you are working on periodic systems, the size of your supercell will not affect much the local electronic distribution, although the program will tend to give slightly different results compared to the unit cell. So your approximation is correct if the atomic environment in the supercell is not modified as specified in the other answer. $\endgroup$– M06-2xCommented Apr 10, 2023 at 13:05
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$\begingroup$ Thank you for your clarification. Yes, for periodic systems, the physics (including a Hubbard potential) is independent of the size of the simulated supercell. Delocalization and more generally correlation effects are not dependent on the size of the simulated supercell in the case of periodic/lightly doped systems. $\endgroup$– ShaCommented Apr 10, 2023 at 16:28