Theory Notes
Contents
Helium Calculations
\( [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} - \frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber \)
Define \(\rho = \frac{Zr}{a_0}\) where \(a_0 = \frac{\hbar^2}{me^2}\) (Bohr radius). Then
\([-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) - Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 + \frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\)
But \(\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}\) is in atomic units (au) of energy. Therefore
\([-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) - \frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi = \varepsilon\psi\nonumber\) where \(\varepsilon = \frac{Ea_0}{Z^2e^2}\)
The problem to be solved is thus \([\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi = \varepsilon\psi\nonumber\)
[figure to be inserted]
The Hartree Fock Method
Assume that \(\psi({\bf r}_1,{\bf r}_2)\) can be written in the form
<math style="horizontal-align:middle;">\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}}[u_1(r_1)u_2(r_2) \pm u_2(r_1)u_1(r_2)]\nonumber</math>
for the \(1S^21S\) ground state
\([-\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi(r_1,r_2) = E\psi(r_1,r_2)\nonumber\)
Substitute into \(<\psi|H-E|\psi>\) and require this expression to be stationary with respect to arbitrary infinitesimal variations \(\delta u_1\) and \(\delta u_2\) in \(u_1\) and \(u_2\). ie
\(\frac{1}{2}<\delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta u_1(r_2)|H-E|u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)>\nonumber\)
\(=\int\delta u_1(r_1)d{\bf r}_1\{\int d{\bf r}_2u_2(r_2)(H-E)[u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)]\}\nonumber\)
\(= 0 \ \ \ for \ arbitrary \ \delta u_1(r_1).\nonumber\)
Therefore \(\{\int d{\bf r}_2 \ldots \} = 0\).
Similarly, the coefficient of \(\delta u_2\) would give
\(\int d{\bf r}_1 u_1(r_1)(H-E)[u_1(r_1)u_2(r_2) \pm u_2(r_1)u_1(r_2)] = 0\nonumber\)
Define
\(I_{12} = \int dru_1(r)u_2(r), \nonumber\)
\(I_{21} = \int dru_1(r)u_2(r), \nonumber\)
\(H_{ij} = \int d{\bf r}u_i(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber\)
\(G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber\)
Then the above equations become the pair of integro-differential equations
\([ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber\)
\([H_0-E+H_{11}+G_{11}(r)]u_2(r) &=& \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber\)
These must be solved self-consistently for the "constants" \(I_{12}\) and \(H_{ij}\) and the function \(G_{ij}(r)\).
The H.F. energy is \(E \simeq -2.87\cdots a.u.\) while the exact energy is \(E = -2.903724\cdots a.u.\)
The difference is called the "correlation energy" because it arises from the way in which the motion of one electron is correlated to the other. The H.F. equations only describe how one electron moves in the average field provided by the other.
Configuration Interaction
Expand \( \psi({\bf r}_1,{\bf r}_2)&=& C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2)\Upsilon^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2)\Upsilon^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm\) exchange where \( \Upsilon^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)&=&\Sigma_{m_1,m_2}\Upsilon^{m_1}_{l_1}({\bf r}_1)\Upsilon^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM> \).
This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to \( ~10^{-7}\) a.u.
Hylleraas Coordinates
<math> [E.A. Hylleraas, Z. Phys. {\bf 48}, 469(1928) and {\bf 54}, 347(1929)] suggested using the co-ordinates $r_1, r_2$ and $r_{12}$ or equivalently \begin{eqnarray} s &=& r_1 + r_2, \nonumber\\ t &=& r_1-r_2, \nonumber\\ u &=& r_{12}\nonumber \end{eqnarray} and writing the trial functions in the form \begin{equation} \Psi({\bf r}_1,{\bf r}_2) = \sum^{1+j+k\leq N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm exchange\nonumber \end{equation} Diagonalizing H in this non-orthogonal basis set is equivalent to solving \begin{equation} \frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber \end{equation} for fixed $\alpha$ and$\beta$.
The diagonalization must be repeated for different values of $\alpha$ and $\beta$ in order to optimize the non-linear parameters.
