I am now reading the paper (or review for beginners), A bird's-eye view of density-functional theory, but I could not understand that the energy minimization problem, in which the derivative of the kinetic energy functional can not be written, can be replaced by solving the Kohn-Sham equation (page 33).

First, the energy minimization problem is written as follows:

\begin{eqnarray} 0 = \frac{\delta E}{\delta \rho(\mathbf r)} = v_\text{ext}(\mathbf r) + v_\text{H}(\mathbf r) + \frac{\delta T_\text{s}[\{\psi_n\}]}{\delta \rho(\mathbf r)} + v_\text{xc}(\mathbf r), \end{eqnarray}

where the external and Hartree terms, $v_\text{ext}$ and $v_\text{H}(\mathbf r)$ are written exactly, and $v_\text{xc}$ can be calculated explicitly once an approximation, such as the local density approximation (LDA) and generalized gradient approximation (GGA), is determined.

Here, because the single-particle exact kinetic energy, $T_\text{s}[\{\psi_n\}]$, is written as an orbital functional (not a density functional $T_\text{s}[\rho]$), we can not write the derivative form.

Subsequently, the paper describes that Kohn and Sham replace this problem by solving an equation (of course, this is the Kohn-Sham equation).

However, there is a gap in my understanding here. I understand the form of the energy minimization problem, and I also understand that the functional derivative of $T_\text{s}[\{\psi_n\}]$ with respect to $\rho$ cannot be written exactly. However, why are these two problems equivalent to solving the KS equation, that is, the eigenvalue problem?


1 Answer 1


The trick is that, instead of solving $\frac{\delta E}{\delta \rho}=0$, we solve the equivalent set of equations $\frac{\delta E}{\delta \psi^*_n}=0, \forall n$ (plus the Lagrange multipliers required to enforce the orthonormalization constraints of the molecular orbitals).

Any energy term $\tilde{E}$ whose functional derivative w.r.t. the density is known explicitly as $\tilde{v}(\mathbf{r})$, gives a contribution of $\tilde{v}(\mathbf{r})\psi_n(\mathbf{r})$ to $\frac{\delta E}{\delta \psi^*_n}$, owing to the chain rule. $T_s = \sum_n\langle \psi_n|-\frac{1}{2}\nabla^2|\psi_n\rangle$ gives a contribution of $-\frac{1}{2}\nabla^2\psi_n(\mathbf{r})$. These explain the LHS of the Kohn-Sham equation, $(-\frac{1}{2}\nabla^2 + v_{\mathrm{ext}}(\mathbf{r}) + v_{\mathrm{H}}(\mathbf{r}) + v_{\mathrm{xc}}(\mathbf{r}))\psi_n(\mathbf{r})$.

The RHS, $\epsilon_n\psi_n(\mathbf{r})$, is due to the Lagrange multipliers which result from the orthonormality of the orbitals. I'll omit the details here because this part of the derivation is exactly the same as the derivation of the Hartree-Fock equation.

  • $\begingroup$ Thank you for always answering my questions so quickly! I now understand this trick and also find KS equation interesting. $\endgroup$
    – neco
    Commented Sep 23, 2021 at 2:25

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