# How is the tight binding model derived from the Kohn-Sham DFT energy?

I am following along a publication titled "Density-functional tight-binding for beginners" (P. Koskinen, V. Mäkinen, Computational Materials Science 2009, 47, 237–253) and I get stomped by a step in their derivation.

The authors start with the following expression for the Kohn-Sham DFT energy: $$E[n]=\sum_{a}f_{a}\langle\psi_{a}|\left(-\frac{1}{2}\,\nabla^{2}+\int V_{\mathrm{ext}}(r)n(r)dr\right)|\psi_{a}\rangle + \frac{1}{2}\int\int\frac{n n^{\prime}}{\left|r-r^{\prime}\right|}dr^{\prime}dr+E_{x c}[n]+E_{I I}$$ where $$f_a \in [0, 2]$$ is the occupation of a single-particle state $$\psi_a$$, $$r$$ is a position vector in 3D, $$n(r)$$ is the function of electron density and $$V_{ext}(r)$$ is an external potential. $$E_{xc}[n]$$ is the exchange-correlation functional which has an unknown form at this point and $$E_{II}$$ is the internuclear repulsion which can be taken as constant.

I believe that the second term in the first parenthesis should instead be just $$V_{ext}(r)$$, but I could be wrong.

The authors then write the following:

Consider a system with density $$n_0(r)$$ that is composed of atomic densities, as if atoms in the system were free and neutral. Hence $$n_0(r)$$ contains (artificially) no charge transfer. The density $$n_0(r)$$ does not minimize the functional $$E[n(r)]$$, but neighbors the true minimizing density $$n_{min}(r) = n_0(r) + \delta n_0(r)$$, where $$\delta n_0(r)$$ is supposed to be small. Expanding $$E[n]$$ at $$n_0(r)$$ to second order in fluctuation $$\delta n(r)$$ the energy reads:

$$E[\delta n]\approx\sum_{a}f_{a}\langle\psi_{a}|-\frac{1}{2}\nabla^{2}+V_{\mathrm{ext}}+V_{H}[n_{0}]+V_{x c}[n_{0}]|\psi_{a}\rangle \\ +\frac{1}{2}\int\int\biggl(\frac{\delta^{2}E_{x c}[n_{0}]}{\delta n\delta n^{\prime}}+\frac{1}{|r-r^{\prime}|}\biggr)\delta n\delta n^{\prime}dr^{\prime}dr-\frac{1}{2}\int V_{H}[n_{0}](r)n_{0}(r)dr \\ +\,E_{x c}[n_{0}]+E_{I I}-\int V_{x c}[n_{0}]n_{0}(r)dr$$

They note that the linear terms in $$\delta n$$ vanish, which I suppose is because we are considering the ground-state density, and thus the derivative of the energy with regards to the density is 0 since we are at a minimum.

The first term and $$E_{II}$$ form the zeroth order term, simple enough. I also managed to derive the second order term (the double integral, the second term). However, I cannot figure out where the three other terms come from and what they mean. Isn't $$E_{xc}[n_0] = \int V_{xc}[n_0]n_0(r)dr$$? And why is there another term for $$E_{H}$$ which is equivalent (?) to the one in the zeroth order term, after some manipulations?

• I haven't had time to look closely, but I think these are the double-counting terms. For example, the <psi_a|V_H|psi)a> term, summed over all states, gives twice the Hartree energy (because every interaction is counted twice); similarly, E_xc is not the eigenvalue sum - try it for the LDA, where the terms are reasonably straightforward. Commented Dec 18, 2022 at 1:24
• I agree that the Hartree energy will be counted twice here, so that could explain one of the term. As for the E_xc, I may be missing some notions here because I don't quite follow what you mean. Also, in the first, exact equation for the energy, E_xc[n] seems to account for the exchange and correlation. It should be also true in the second equation, no? Commented Dec 19, 2022 at 2:11
• I found a detailed derivation in W. M. C. Foulkes, R. Haydock, Phys. Rev. B 1989, 39, 12520–12536. The additional terms are not for double-counting but in fact related to the condition that the density is a ground-state density. I will post a more detailed answer once I'm done going over the derivation. Commented Dec 20, 2022 at 13:26

## 1 Answer

The additional terms come from the Kohn-Sham equations. Everything is detailed quite nicely in W. M. C. Foulkes, R. Haydock, Phys. Rev. B 1989, 39, 12520–12536. Here is a summary of mine:

We start by considering a fictitious system of non-interacting electrons with the same groundstate density $$n(r)$$ as the real system. From the Schrödinger equation, we know the energy of each electron in the fictitious system: $$\left( \frac{1}{2}\nabla^2 + V(r) \right)\psi_i(r) = \epsilon_i \psi_i(r)$$ where $$V(r)$$ is the potential which yields the groundstate density $$n(r)$$ in this non-interacting system. The total energy is then simply the sum of one-electron energies $$\epsilon_i$$. We now consider this case from a DFT perspective: $$E[n] = T_s[n] + F[n]$$ where $$T_s[n]$$ is the kinetic energy functional for non-interacting electrons and $$F[n]$$ is the energy associated with the potential $$V(r)$$. In the first equation, we can left-multiply both sides by $$\psi_i^*(r)$$ and derive the following relationship:

$$\int \psi_i^*(r)\frac{1}{2}\nabla^2\psi_id^3r = T_s[n] = \sum_{i=1}^N\epsilon_i - \int V(r)n(r) d^3r$$

Then, we consider the condition that we want a groundstate density, meaning that $$\delta E[n]/\delta n(r) = 0$$. Starting from the second equation:

\begin{align} \int \frac{\delta E[n]}{\delta n(r)}\delta n(r) d^3r &= \int\frac{\delta T_s[n]}{\delta n(r)}\delta n(r) d^3r + \int\frac{\delta F[n]}{\delta n(r)}\delta n(r) d^3r \approx 0 \\ &= \int \left(-V(r) + \frac{\delta F[n]}{\delta n(r)}\right)\delta n(r) d^3r \approx 0 \\ \end{align}

We conclude that $$V(r) = \delta F[n]/\delta n(r) = V_H[n](r) + V_{ext}(r) + \mu_{xc}[n](r)$$ for this to be true. Putting it all together:

\begin{align} E_0[n_0] &= \sum_{i=1}^N\epsilon_i - \int V(r)n_0(r)d^3r + F[n_0(r)] \\ &= \sum_{i=1}^N\epsilon_i - \int \left.\frac{\delta F}{\delta n}\right|_{n_0(r)} n_0(r)d^3r + F[n_0(r)] \\ &= \sum_{i=1}^N\epsilon_i - \int \left(V_{ext}(r) + V_H[n_0(r)] + \mu_{xc}[n_0(r)]\right)n_0(r)d^3r + E_{ext}[n_0(r)] + E_H[n_0(r)] + E_{xc}[n_0(r)] \\ &= \sum_{i=1}^N\epsilon_i - \int V_H[n_0(r)]n_0(r)d^3r - \int\mu_{xc}[n_0(r)]n_0(r)d^3r + E_H[n_0(r)] + E_{xc}[n_0(r)] \\ &= \sum_{i=1}^N\epsilon_i - 2E_H[n_0(r)] - \int\mu_{xc}[n_0(r)]n_0(r)d^3r + E_H[n_0(r)] + E_{xc}[n_0(r)] \\ &= \sum_{i=1}^N\epsilon_i - E_H[n_0(r)] - \int\mu_{xc}[n_0(r)]n_0(r)d^3r + E_{xc}[n_0(r)] \\ \end{align}

And this is the true zeroth order term in the functional expansion. The second order terms are added to this to yield the final expression.

• +10. Beautiful first answer! Welcome to our new community and thank you for your contributions. We hope to see much more of you in the future! Commented Dec 21, 2022 at 21:49