I am reading introductory DFT (Density Functional Theory) and I came across this, enter image description here

Can someone explain why the system needs to be neutral for energy to be finite? and What does organising in neutral groups mean and has relevance for calculation??

[Edit]: Just converting the relevant part into latex,

  • Any extended system must be neutral if the energy is to be finite
  • Terms in the energy must be organize in neutral groups for actual evaluation $$E^{CC} = E_{Hartree} + \int d^3rV_{ext}(\vec{r})n(\vec{r}) + E_{11}$$ where $E_{Hartree} = \frac{1}{2} \int d^3 r d^3r'\frac{n(r)n(r')}{|r-r'|}$ is the self interaction energy of density n(r). Since the second term is the inetraction between electrons and nuclei and the $E_{11}$ is the interaction between nuclei, the whole equation is neutral grouping of terms

[Book: Richard Martin, Electronic Structures: Basic Theory and Practice Methods p64]

  • $\begingroup$ Please give a detailed reference (book, author, edition, chapter, page...) And please use MathJax/the quote function instead of images. $\endgroup$ Commented Jun 16, 2023 at 6:50
  • $\begingroup$ @TobiasFünke I've made the edits $\endgroup$ Commented Jun 16, 2023 at 6:59

1 Answer 1


In a homogeneous charged system with a charge density $\rho_Q$, every finite volume $\Delta V$ has a charge $Q= \rho_Q \Delta V$. The infinite volume will have an infinite charge, and the potential energy associated with such infinite charge, due to the long-range nature of Coulomb interaction, will diverge faster than the volume (it is a simple exercise to evaluate how the potential energy of a uniformly charged sphere grows with the volume). Therefore, there is no way to get a finite density of energy.

Even in the case of a globally neutral system with positive and negative charges, one has to sum different spatial contributions to the energy with some care. In particular, due to the long-range character of Coulomb's interaction, it turns out that the sum of all the local contributions to the potential energy is conditionally convergent. Depending on the order of summation, one could get any possible real number, including $ \pm \infty$.

Therefore, one needs a globally neutral system and a proper summation prescription to get a meaningful energy density.

  • $\begingroup$ But our volume is usually finite right? I'm doing DFT using quantum espresso and from what I understand I can limit the region where this is being applied $\endgroup$ Commented Jun 16, 2023 at 13:53
  • $\begingroup$ @HarshdeepChhabra The fact that you work on a few cells of an infinite lattice does not modify the situation. If you allow even a small charge, the infinite perfect crystal will get an infinite charge and, even worst, an infinite energy per cell. Sometimes one performs calculations with an excess of ions or electrons, but in those cases a uniform neutralizing background is added. $\endgroup$ Commented Jun 16, 2023 at 14:14
  • $\begingroup$ so when lattice has a charge or ions are there how is those problem accounted? Do they do modify boundary conditions and add counterions? $\endgroup$ Commented Jun 16, 2023 at 14:16
  • $\begingroup$ In some cases, one adds counterions, otherwise, a neutralizing background is introduced. As far as I remember (but it has been a long time since I used QE, therefore, do not trust me without checking directly), the neutralizing background is automatically introduced if the simulation cell is not neutral. $\endgroup$ Commented Jun 16, 2023 at 14:27

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