2.52: The SCF Method for Two Electrons Using a Gaussian Wave Function

Gaussian Trial Wave Function:

\[ \Psi (r,~ \beta ) = \left( \frac{2 \beta}{ \pi} \right)^{ \frac{3}{4}} \text{exp} \left( - \beta r^2 \right) \nonumber \]

Calculate kinetic energy:

\[ \begin{array}{c|c} \int_0^{ \infty} \Psi (r,~ \beta ) \left[ - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta)) \right] 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{3}{2} \beta \end{array} \nonumber \]

Calculate electron‐nucleus potential energy:

a. Calculate the electric potential of one of the electrons in the presence of the other:

\[ \begin{array}{c|c} \frac{1}{r} \int_0^r \Psi (x,~ \beta )^2 4 \pi x^2 dx + \int_r^{ \infty} \frac{ \Psi (x,~ \beta)^2 4 \pi x^2}{x} dx & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{ \text{erf} \left( r 2^{ \frac{1}{2}} \beta^{ \frac{1}{2}} \right)}{r} \end{array} \nonumber \]

b. Calculate the electron-electron potential energy using the result of part a:

\[ \begin{array} \int_0^{ \infty} \Psi (r,~ \beta )^2 \left( \frac{ \text{erf} \left( r 2^{ \frac{1}{2}} 2^{ \frac{1}{2}} \right)}{r} \right) 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{2}{ \pi} ( \beta \pi )^{ \frac{1}{2}} \end{array} \nonumber \]

SCF Calculation

1. Supply nuclear charge and an input value for β: \[ \begin{matrix} Z = 2 & \beta = 0.767 & \alpha = Z \end{matrix} \nonumber \]
2. Define orbital energies of the electrons in terms of the variational parameters: \[ \begin{matrix} \text{Orbital energy of the } \alpha \text{ electron:} & \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = \frac{3 \alpha}{2} - Z \sqrt{ \frac{8 \alpha}{ \pi}} + \sqrt{ \frac{8 \alpha \beta}{ \pi ( \alpha + \beta)}} \\ \text{Orbital energy of the } \beta \text{ electron:} & \varepsilon_{1s \beta} ( \alpha,~ \beta ) = \frac{3 \beta}{2} - Z \sqrt{ \frac{8 \beta}{ \pi}} + \sqrt{ \frac{8 \alpha \beta}{ \pi ( \alpha + \beta)}} \end{matrix} \nonumber \]
3. Minimize orbital energies with respect to α and β: \[ \begin{matrix} \text{Given} & \frac{d}{d \alpha} \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = 0 & \alpha = \text{Find} ( \alpha ) & \alpha = 0.7670 & \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = -0.6564 \\ \text{Given} & \frac{d}{d \beta} \varepsilon_{1s \beta} ( \alpha,~ \beta ) = 0 & \beta = \text{Find} ( \beta ) & \beta = 0.7670 & \varepsilon_{1s \beta} ( \alpha,~ \beta ) = -0.6564 \end{matrix} \nonumber \]
4. Calculate the energy of the atom: \[ \begin{matrix} E_{atom} = \frac{3 \alpha}{2} & \frac{3 \beta}{2} - Z \sqrt{ \frac{8 \alpha}{ \pi}} - Z \sqrt{ \frac{8 \beta}{ \pi}} + \sqrt{ \frac{8 \alpha \beta}{ \pi ( \alpha + \beta)}} & E_{atom} = -2.3010 \end{matrix} \nonumber \]
5. Record results of the SCF cycle and return to step 1 with the new and improved input value for β.
6. Continue until self‐consistency is achieved.
7. Verify the results shown below for He. Repeat for Li⁺, Be²⁺ and B³⁺.

\[ \begin{pmatrix} \beta \text{(input)} & \alpha & \varepsilon_{1s \alpha} & \beta & \varepsilon_{1s \beta} & E_{atom} \\ 2.000 & 0.4514 & -0.4988 & 0.9303 & -0.8031 & -2.2703 \\ 0.9303 & 0.6943 & -0.6117 & 0.8023 & -0.6816 & -2.2996 \\ 0.8023 & 0.7504 & -0.6454 & 0.7749 & -0.6618 & -2.3009 \\ 0.7749 & 0.7633 & -0.6539 & 0.7688 & -0.6576 & -2.3010 \\ 0.7688 & 0.7661 & -0.6588 & 0.7674 & -0.6567 & -2.3010 \\ 0.7674 & 0.7668 & -0.6563 & 0.7671 & -0.6564 & -2.3010 \\ 0.7671 & 0.7669 & -0.6564 & 0.7670 & -0.6564 & -2.3010 \\ 0.7670 & 0.7670 & -0.6564 & 0.7670 & -0.6564 & -2.3010 \end{pmatrix} \nonumber \]