Theory Notes

From Dr. GWF Drake's Research Group
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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.

Hylleras Co-ordinates

Completeness

Solutions of the Eigenvalue Problem

Brute Force Method

The Power Method

Matrix Elements of H

Radial Integrals and Recursion Relations

The Radial Recursion Relation

The General Integral

Graphical Representation

[figure to be inserted]

Matrix Elements of H

Problem

General Hermitean Property

Optimization of Non-linear Parameters

  • Difficulties
  • Cure

The Screened Hyrdogenic Term

Small Corrections

  • Mass Polarization

Test

\(\mid \psi_P_A^0 \rangle = \mid \bf{P_A} \rangle \mid 0 \rangle \)