Theory Notes

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 \([\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi = \varepsilon\psi\nonumber\)

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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 $r_1, 4_2$ 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.<math>

Completeness

<math> The completeness of the above basis set can be shown by first writing $r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\Theta_{12})$ and $\cos(\Theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_l(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)$ consider first S-states. The $r_{12}^0$ terms are like the ss terms in a CI calculation. The $r_{12}^2$ terms bring in p-p type contributions, and the higher powers bring in d-d, f-f etc type terms. In general \begin{equation} P_l(\cos(\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber \end{equation}

For P-states, one would have similarly \begin{eqnarray} &r_{12}^0&\ \ \ \ \ \ \ \ (sp)P\nonumber\\ &r_{12}^2&\ \ \ \ \ \ \ \ (pd)P\nonumber\\ &r_{12}^4&\ \ \ \ \ \ \ \ (df)P\nonumber\\ &\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\nonumber \end{eqnarray}

For D-states \begin{eqnarray} &r_{12}^0&\ \ \ \ \ \ \ \ (sp)D\ \ \ \ \ \ \ \ (pp^\prime)D\nonumber\\ &r_{12}^2&\ \ \ \ \ \ \ \ (pd)D\ \ \ \ \ \ \ \ (dd^\prime)D\nonumber\\ &r_{12}^4&\ \ \ \ \ \ \ \ (df)D\ \ \ \ \ \ \ \ (ff^\prime)D\nonumber\\ &\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\nonumber\end{eqnarray}

In this case, since there are two ``lowest-order couplings to form a D-state, both must be present in the basis set. ie

\begin{eqnarray} \Psi(r_2,r_2) &=& \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta r-2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)\nonumber\\ &+&\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12}e^{-\alpha^\prime r_1 - \beta^\prime r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber \end{eqnarray}

For F-states, one would need $(sf)F$ and $(pd)F$ terms.

For G-states, one would need $(sg)G$, $(pf)G$ and $(dd^\prime)G$ terms.

Completeness of the radial functions can be proven by considering the Stern-Liouville problem

\begin{equation} \left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber \end{equation} or \begin{equation} \left(-\frac{1}{2}\frac{1}{r^2}\left({r^2}\frac{\partial}{\partial r}\right) - \frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber \end{equation} for fixed E and variable $\lambda$ (nuclear charge).

The eigenvalues are $\lambda_n = (E/E_n)^{1/2}$, where $E_n =- \frac{1}{2n^2}$

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\begin{equation} u_{nl}(r) = \frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha r},\nonumber \end{equation} with $\alpha = (-2E)^{1/2}$ and $n\geq l+1$.

Unlike the hydrogen spectrum, which has both a discrete part for $E<0$ and a continuous part for $E>0$, this forms an entirely discrete set of finite polynomials, called Sturmian functions. They are orthogonal with respect to the potential -ie \begin{equation} \int^\infty_0 r^2dru_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) = \delta_{n,n^\prime}\nonumber \end{equation}

Since they become complete in the limit $n\rightarrow\infty$, this assures the completeness of the variational basis set.

[see B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) {\bf 44}, 9 and 27 (1977)].<math>

Graphical Representation

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Optimization of Non-linear Parameters

• Difficulties
• Cure

Small Corrections

• Mass Polarization