Density Functional Theory: Conceptual Overview

14.1 The Hohenberg—Kohn Theorems

Theorem 1: The ground-state electron density $n(\mathbf{r})$ uniquely determines the external potential $V_{\text{ext}(\mathbf{r})}$ (up to an additive constant), and hence the full many-body Hamiltonian and all ground-state properties.

Theorem 2: The ground-state energy is a functional of the density: $E[n] = F_{\text{HK}[n] + \int V_{\text{ext}(\mathbf{r})n(\mathbf{r})\,d^3r}}$And the variational principle applies: $E_0 \leq E[n]$ for any trial density $n(\mathbf{r})$.

14.2 Kohn—Sham Equations

The interacting system is mapped to a fictitious system of non-interacting electrons in an effective potential:

\left[-\frac{\hbar^2}{2m}\nabla^2 + V_{\text{eff}[n](\mathbf{r})\right]\psi_i(\mathbf{r}) = \varepsilon_i\psi_i(\mathbf{r})}

$n(\mathbf{r}) = \sum_{i=1}^{N}|\psi_i(\mathbf{r})|^2 \quad \text{(summing over occupied states)}$

$V_{\text{eff} = V_{\text{ext} + V_H[n] + V_{\text{xc}[n]}}}$

$V_H[n](\mathbf{r}) = e^2\int\frac{n(\mathbf{r}")}{|\mathbf{r} - \mathbf{r}'|}\,d^3r' \quad \text{(Hartree potential)}$

The exchange-correlation functional $V_{\text{xc}[n]}$ contains all many-body effects beyond the classical Hartree approximation.

14.3 Self-Interaction Error

The Hartree potential includes the interaction of each electron with itself. This self-interaction error is not cancelled by the local density approximation (LDA) for $V_{\text{xc}}$. Consequences:

• Wrong asymptotic behaviour: $V_{\text{eff}(r \to \infty) \to -e^2/r}$ (correct) vs. $V_{\text{eff} \to 0}$ (LDA, wrong)
• Underestimation of band gaps by 30—50%
• Incorrect description of charge transfer excitations

Hybrid functionals (e.g., B3LYP, HSE06) and range-separated functionals partially correct this.

Worked Examples

Example 1: Infinite square well

Problem. Find the energy levels and normalised wave functions for a particle in a 1D infinite square well of width $a$.

Solution. $\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$, $E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}$, $n = 1, 2, 3, \ldots$

$\blacksquare$

Example 2: Uncertainty principle

Problem. An electron is confined to a region of width $0.1 \mathrm{ nm}$. Estimate the minimum uncertainty in its momentum.

Solution. $\Delta x = 0.1 \times 10^{-9} \mathrm{ m}$. ${\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2 \times 10^{-10}} = 5.28 \times 10^{-25} \mathrm{ kg\cdot} m/s}$.

$\blacksquare$

Common Pitfalls

• Confusing the wave function and probability density. $|\psi(x)|^2$ is the probability density; $\psi$ itself is complex and not directly observable. Fix: $P(x)\, dx = |\psi(x)|^2\, dx$ is the probability of finding the particle in $[x, x + dx]$.
• Wrong commutator interpretation. If $[\hat{A}, \hat{B}] = 0$, the observables share eigenstates and can be simultaneously measured. If not, they obey the uncertainty principle. Fix: $[\hat{x}, \hat{p}] = i\hbar$ implies $\Delta x \cdot \Delta p \geq \hbar/2$.
• Confusing time-dependent and time-independent Schrödinger equations. Time-dependent: $i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi$. Time-independent: $\hat{H}\phi = E\phi$. Fix: Time-independent gives stationary states (energy eigenvalues); time-dependent describes evolution.

Summary

• Postulates: state vector $|\psi\rangle$, observables as Hermitian operators, measurement gives eigenvalues.
• Schrödinger equation: $i\hbar\partial\psi/\partial t = \hat{H}\psi$; stationary states: $\hat{H}\phi_n = E_n\phi_n$.
• Commutators and uncertainty: $[\hat{A}, \hat{B}] \neq 0 \Rightarrow \Delta A \cdot \Delta B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|$.
• Key systems: infinite square well, harmonic oscillator, hydrogen atom.

Cross-References