# 13.1: Prelude to Thomas-Fermi Analysis

Where are the electrons found around an atom? This question is difficult for a few reasons. First, at temperatures above absolute zero, electrons are in continual motion. Second, the Heisenberg uncertainty principle tells us that we can never know the position and momentum of electrons simultaneously with complete accuracy. However, this question isn't hopeless. We can find the charge density $$\rho_{ch}$$ which tells us, statistically on average, where the electrons are most likely to be found. Understanding the distribution of electrons in a material is vital to understanding the chemical properties, such as the strength of chemical bonds, as well as the electrical properties, such as how much energy is required to remove electrons. To answer this question, we will use calculus of variations. The generalized path will be voltage $$V$$ , and the generalized potential will be charge density $$\rho_{ch}$$. A Lagrangian describes an energy difference, and the Lagrangian will have the form

$\mathcal{L}=\mathcal{L}\left(r, V, \frac{d V}{d r}\right).$

The path found in nature is the one that minimizes the action.

$\delta \int_{r_{1}}^{r_{2}} \mathcal{L} d r=0$

In this problem, the independent variable is position, not time. We will set up the Euler-Lagrange equation then solve it to find the equation of motion.

Most of this chapter consists of a derivation of the resulting equation of motion called the Thomas-Fermi equation. With a bit of algebra, we can find both the voltage and the charge density around the atom from the solution to the Thomas-Fermi equation. The procedure is as follows.

• Describe the first form of energy, $$E_{Coulomb\, e \,nucl} + E_{e \,e \,interact}$$, in terms of path $$V$$ . The resulting energy density is $\frac{E_{Coulomb \,e\, nucl}}{\mathbb{V}}+\frac{E_{e \,e \,interact}}{\mathbb{V}}=\frac{1}{2} \epsilon|\overrightarrow{\nabla} V|^{2}$ where $$\epsilon$$ represents permittivity and $$\mathbb{V}$$ represents volume.
• Describe the second form of energy $$E_{kinetic\, e}$$ in terms of path $$V$$. The resulting energy density is $\frac{E_{kinetic \, e}}{\mathbb{V}}=c_{0} V^{5 / 2}$ where $$c_0$$ is a constant. This step will require the idea of reciprocal space.
• Write down the Hamiltonian $$H \left(r, V, \frac{dV}{dr} \right)$$ and Lagrangian $$\mathcal{L} \left(r, V, \frac{dV}{dr} \right)$$.
• Set up the Euler-Lagrange equation. $\frac{\partial \mathcal{L}}{\partial V}-\overrightarrow{\nabla} \cdot\left(\frac{\partial \mathcal{L}}{\partial\left(\frac{d V}{d r}\right)}\right) \hat{a}_{r}=0$
• Solve the Euler-Lagrange equation for the equation of motion. The result is $\frac{5}{2} c_{0} V^{3 / 2}-\epsilon \nabla^{2} V=0.$
• Change variables to clean up the equation of motion. The resulting equation is called the Thomas-Fermi equation. $\frac{d^{2} y}{d t^{2}}=\mathrm{t}^{-1 / 2} y^{3 / 2}$
• Voltage and charge density are algebraically related to the quantity $$y$$ in the equation above.

To attempt to find charge density and voltage as a function of position $$r$$ from the center of the atom, we will have to make some rather drastic assumptions. This analysis follows works of Thomas [173] and Fermi [174] which were originally completed around 1927. This derivation is discussed by numerous other authors as well [6] [46] [136] [175]. Because of the severe assumptions made below, the results will not be very accurate. However, more accurate numerical calculations are based on improved versions of the techniques established by Thomas and Fermi. We are discussing the most simplified version of the derivation, but this is the basis of more accurate approaches.