# 418: Energy Minimization - Four Methods Using Mathcad

Using $$\psi ( \alpha , r) = \frac{ \alpha ^{3}}{ \pi} exp( - \alpha r)$$ as a trial wave function for the helium atom electrons leads to the following energy expression in terms of the variational parameter, α.

$$E( \alpha ) = \alpha ^{2} - 4 \alpha + \frac{5}{8} \alpha$$

The first term is electron kinetic energy, the second electron‐nucleus potential energy and the final term electron‐electron potential energy.

Mathcad provides four methods for energy minimization with respect to $$α$$. The second and third methods require a seed value for $$α$$.

## First method

$\alpha = \frac{d}{d \alpha} E( \alpha ) = 0 |_{float, 5}^{solve, \alpha} \rightarrow 1.6875$

$$E ( \alpha ) = -2.8477$$

## Second method

α = 1. Given $$\frac{d}{d \alpha} E ( \alpha ) = 0$$ $$\alpha = Find ( \alpha )$$ α = 1.685 $$E ( \alpha ) = -2.8477$$

## Third method

α = 1. α : = Minimize(E, α) α = 1.685 $$E ( \alpha ) = -2.8477$$

## Fourth method

Clear memory of α and X: α = α Z = Z

$$En ( \alpha ,~Z) = \alpha ^{2} - 2 Z \alpha + \frac{5}{8} \alpha$$ $$\frac{d}{d \alpha} En( \alpha ,~Z) = 0~solve, \alpha \rightarrow Z - \frac{5}{16}$$

$$En ( \alpha ,~Z) = \alpha ^{2} - 2 Z \alpha + \frac{5}{8}~substitute, \alpha = Z - \frac{5}{16} \rightarrow - \frac{(16 Z - 5)^{2}}{256}$$

$$En ( \alpha ,~2) = -2.8477$$ $$En \alpha ,~3) = -7.2227$$ $$En( \alpha ,~4) = -13.5977$$

Two variables: a molecular orbital calculation yields the following result for the energy of the hydrogen molecule ion as a function of the internuclear separation and the orbital decay constant.

$$1s_{a} = \frac{ \alpha ^{3}}{ \pi} exp(- \alpha r_{a})$$

$$1s_{b} = \frac{ \alpha ^{3}}{ \pi} exp(- \alpha r_{b})$$

$$S_{ab} = \int 1s_{a} 1s_{b} d \tau$$

$$\psi_{mo} = \frac{1s_{a} + 1s_{b}}{ \sqrt{2+2 S_{ab}}}$$

$E( \alpha, R) = \frac{- \alpha^{2}}{2} + \frac{[ \alpha^{2} - \alpha - \frac{1}{R} + \frac{1 + \alpha R}{R} exp(-2 \alpha R) + \alpha ( \alpha - 2) (1+ \alpha R) exp(- \alpha R)]}{[1 + exp(- \alpha R) (1+ \alpha R + \frac{ \alpha^{2} R^{3}}{3})]} + \frac{1}{R}$

α = 1 R = 1 $$\begin{pmatrix} \alpha \\ R \end{pmatrix}$$ = Minimize(E, α, R) $$\begin{pmatrix} \alpha \\ R \end{pmatrix}$$ = $$\begin{pmatrix} 1.2380 \\ 2.0033 \end{pmatrix}$$ E(α, R) = -0.5865

α = 1 Energy = -2 Given Energy = E(α, R) $$\frac{d}{d \alpha}$$ E(α, R) = 0 Energy(R) = Find(α, Energy)

R = .2, .25 .. 10 T(R) = -Energy(R)1 - R $$\frac{d}{dR}$$ Energy(R)1 V(R) = 2  Energy(R)1 + R $$\frac{d}{dR}$$ Energy(R)1