2.51: Outline of the SCF Method for Two Electrons
- Page ID
- 158524
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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+
\[ \begin{pmatrix} \beta \text{ (input)} & \alpha & \varepsilon_{1s \alpha} & \beta & \varepsilon_{1s \beta} & E_{atom} \\ 2.000 & 1.5999 & -0.8116 & 1.7126 & -0.9250 & -2.8449 \\ 1.7126 & 1.6803 & -0.8887 & 1.6895 & -0.8987 & -2.8476 \\ 1.6895 & 1.6869 & -0.8959 & 1.6877 & -0.8967 & -2.8477 \\ 1.6877 & 1.6874 & -0.8964 & 1.6875 & -0.8965 & -2.8477 \\ 1.6875 & 1.6875 & -0.8965 & 1.6875 & -0.8965 & -2.8477 \end{pmatrix} \nonumber \]