2.51: Outline of the SCF Method for Two Electrons
- Page ID
- 158524
Trial Wave Function:
\[ \Psi (r,~ \beta ) = \sqrt{ \frac{ \beta^3}{ \pi}} \text{exp} ( - \beta r) \nonumber \]
Calculate kinetic energy:
\[ \begin{array}{c|c} T_e ( \beta ) = \int_0^{ \infty} \Psi (r,~ \beta ) \left[ - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta )) \right] 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow (- \beta ) Z \end{array} \nonumber \]
Calculate electron‐nucleus potential energy:
\[ \begin{array}{c|c} V_{ne} ( \beta,~ Z) = \int_0^{ \infty} \Psi (r,~ \beta ) \frac{-Z}{r} 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow (- \beta ) Z \end{array} \nonumber \]
Calculation of electron‐electron potential energy:
a. Calculate the electrostatic potential due to the β electron:
\[ \begin{array}{c c |c} \Phi ( \beta,~ r) = & \frac{1}{r} \int_0^{ \tau} \Psi (x,~ \beta)^2 4 \pi x^2 dx ... & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{- \left[ e^{(-2)r \beta} \beta r + e^{(-2) r \beta} - 1 \right]}{r} \\ & + \int_r^{ \infty} \frac{ \Psi (x,~ \beta )^2 4 \pi x^2 dx}{x} dx \end{array} \nonumber \]
b. Calculate the electron‐electron potential energy of the α and β electrons using result of part a:
\[ \begin{array}{c|c} V_{ee} ( \alpha,~ \beta ) = \int_0^{ \infty} \Psi (r,~ \alpha )^2 \Phi ( \beta,~ r) 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0,~ \alpha >0} \rightarrow \alpha \beta \frac{ \alpha^2 + 3 \alpha \beta + \beta^2}{ \left( \alpha^2 + 2 \beta \alpha + \beta^2 \right) ( \beta + \alpha )} \end{array} \nonumber \]
SCF Calculation
- Supply nuclear charge and an input value for β: \[ \begin{matrix} Z = 2 & \beta = 2.0 & \alpha = Z \end{matrix} \nonumber \]
- Define orbital energies of the electrons in terms of the variational parameters: \[ \begin{matrix} \text{Orbital energy of the } \alpha \text{ electron:} & \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = T_e ( \alpha ) + V_{ne} ( \alpha,~ Z) + V_{ee} ( \alpha,~ \beta ) \\ \text{Orbital energy of the } \beta \text{ electron:} & \varepsilon_{1s \beta} ( \alpha,~ \beta ) = T_e ( \beta ) + V_{ne} ( \beta,~ Z) + V_{ee} ( \alpha,~ \beta ) \end{matrix} \nonumber \]
- Minimize orbital energies with respect to α and β: \[ \begin{matrix} \text{Given} & \frac{d}{d \alpha} \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = 0 & \alpha = \text{Find} ( \alpha ) & \alpha = 1.5999 & \varepsilon_{1s \alpha} ( \alpha,~ \beta ) = -0.8116 \\ \text{Given} & \frac{d}{d \beta} \varepsilon_{1s \beta} ( \alpha,~ \beta ) = 0 & \beta = \text{Find} ( \beta ) & \beta = 1.7126 & \varepsilon_{1s \beta} ( \alpha,~ \beta ) = -0.9250 \end{matrix} \nonumber \]
- Calculate the energy of the atom: \[ \begin{matrix} E_{atom} = T_e ( \alpha ) +V_{ne} ( \alpha,~ Z) + T_e ( \beta ) + V_{ne} ( \beta,~Z) + V_{ee} ( \alpha,~ \beta ) & E_{atom} = -2.8449 \end{matrix} \nonumber \]
- Record results of the SCF cycle and return to step 1 with the new and improved input value for β.
- Continue until self‐consistency is achieved.
- Verify the results shown below for He. Repeat for Li+, Be2+ and B3+
\[ \begin{pmatrix} \beta \text{ (input)} & \alpha & \varepsilon_{1s \alpha} & \beta & \varepsilon_{1s \beta} & E_{atom} \\ 2.000 & 1.5999 & -0.8116 & 1.7126 & -0.9250 & -2.8449 \\ 1.7126 & 1.6803 & -0.8887 & 1.6895 & -0.8987 & -2.8476 \\ 1.6895 & 1.6869 & -0.8959 & 1.6877 & -0.8967 & -2.8477 \\ 1.6877 & 1.6874 & -0.8964 & 1.6875 & -0.8965 & -2.8477 \\ 1.6875 & 1.6875 & -0.8965 & 1.6875 & -0.8965 & -2.8477 \end{pmatrix} \nonumber \]