Solution of the Hartree-Fock-Roothaan equations for helium

To solve the Hartree-Fock-Roothaan equations for He, we begin by noting that $n'=2$ in the spin orbitals $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ and $\phi_i(x)=\sum_{\beta=1}^{n'}c_{i\beta}\chi_{\beta}(x)$ . If the basis wavefunctions $\chi_{\alpha}(x)$ and $\chi_{\beta}(x)$ are real, $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ becomes $\phi_i(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}\chi_{\alpha}(x)$ . With reference to eq157, we have

$(H_{11}-\lambda_iS_{11})c_{i1}+(H_{12}-\lambda_iS_{12})c_{i2}=0\;\;\;\;when\;\alpha=1$

$(H_{21}-\lambda_iS_{21})c_{i1}+(H_{22}-\lambda_iS_{22})c_{i2}=0\;\;\;\;when\;\alpha=2$

where $H_{\alpha\beta}=\int\chi_{\alpha}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_{\beta}(x)dx+\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_{\alpha}(x)\chi_{\beta}(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$ (assuming no contribution from the exchange integral) and $S_{\alpha\beta}=\int\chi_{\alpha}(x)\chi_{\beta}(x)dx$ .

The pair of simultaneous equations can be represented by the following matrix equation:

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{21}-\lambda_iS_{21} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(158a)$

Let the basis wavefunctions be $\chi_1=\sqrt{\frac{\zeta_1^3}{\pi}}e^{-\zeta_1r}$ and $\chi_x=\sqrt{\frac{\zeta_x^3}{\pi}}e^{-\zeta_xr}$ . Since these Slater-type orbitals are real, $S_{12}=S_{21}$ . Eq158 becomes

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{21}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(159)$

Question

Show that $H_{12}=H_{21}$ .

Answer

Since $\chi_1$ and $\chi_2$ are real,

$H_{12}=\int\chi_1(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_2(x)dx+$ $\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_1(x)\chi_2(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$
Using the Hermitian property of the KE operator,

$H_{12}=\int\chi_2(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_1(x)dx+$ $\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_2(x)\chi_1(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'=H_{21}$

Therefore, eq159 becomes

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{12}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(160)$

Eq160 is a linear homogeneous equation, which has non-trivial solutions if

$\begin{vmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{12}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{vmatrix}=0$

Expanding the determinant and noting that $S_{ii}=1$ , we get the characteristic equation:

$(1-S_{12}^2)\lambda_i^2+(2H_{12}S_{11}-H_{11}-H_{22})\lambda_i+H_{11}H_{22}-H_{12}^2 =0\;\;\;\;\;\;\;\;(161)$

The computation process is as follows:

Evaluate the integrals $H_{11}$ , $H_{12}$ , $H_{22}$ and $S_{12}$ in eq161 either analytically or numerically by letting $\zeta_1=1.6$ and $\zeta_2=2.0$ , and using the initial guess values of $c_{21}=c_{22}=0.50$ .
Substitute the evaluated integrals in eq161, solve for $\lambda_1$ and retain the lower root. Substitute $\lambda_1$ back in eq159 to obtain an expression of $c_{11}$ in terms of $c_{12}$ .
Use the expression derived in part 2, together with $\int\phi_i^*(x)\phi_i(x)d\tau=c_{11}^2S_{11}+2c_{11}c_{12}S_{12}+c_{22}^2S_{22}=1$ , to solve for $c_{11}$ and $c_{12}$ , which are then used as improved values of $c_{21}$ and $c_{22}$ for the next iteration.
Repeat steps 1 through 4 until $c_{11}$ , $c_{12}$ and $\lambda_1$ are invariant up to 6 decimal places.